ANSWER:
A tangent line touches a circle at a single point.
STEP-BY-STEP EXPLANATION:
A tangent to a circle is a straight line that touches the circle at only one point. This point is called the point of tangency.
Statement: Using the figure, determine the following ratios. ∆RTS is a right triangle with a right angle at
By definition:
[tex]\text{tan(angle)}=\frac{\text{opposite side}}{adjacent\text{ side}}[/tex]From the picture:
[tex]\begin{gathered} \tan (S)=\frac{80}{18}=4.4444 \\ \tan (R)=\frac{18}{80}=0.225 \end{gathered}[/tex]Given that lines b and care parallel, select all that apply.Which angles are congruent to 3?d123/4→b5/67/80 240270_226
We want to look which angles are congruent to 3 let's analyze the possible options:
Angle 4 is ot congruent since it's not a vertical angle respect to 3
Angle 7 is congruent since 7 and 3 are corresponding angles
Angle 2 is congruent since 2 and 3 are opposite angles
Angle 6 is congruent since 6 and 3 are consecutive interior angles
So
John and Amber work at an ice cream shop. The hours worked and wages earned are given for each person.
John's wages are proportional to time if every time that we divide wages by time, we get a constant, so we can make the following table:
Time Wages Wages/Time (y/x)
2 18 18/2 = 9
3 27 27/3 = 9
4 36 36/4 = 9
Since Wages/Time is always equal to 9, John's wages are proportional to time.
Answer: John's wages are proportional to time.
61.2 + 2x - 10 = -* - 71 + 4x
61.2 + 2x - 10 = - * - 71 + 4x
What do I have to do?
Milanputting money into a savings account. He starts with $350 in the savings account, and each week he adds $60.Let S represent the total amount of money in the savings account (in dollars), and let W represent the number of weeks Milan has been adding money. Write anequation relating S to W. Then use this equation to find the total amount of money in the savings account after 17 weeks.
We know that he starts with $650 in the savings account, and each week he adds $30.
If S represents the total amount of money and w the number of weeks, an equation that realtes S to w is compounded by:
1. An constant value that is the value which he starts
S = 650
2. Multiplying the number of weeks by the money he adds weekly
S = 650 + w30
So, the equation that relates S to w is
[tex]S=650+w30[/tex]Now, we must calculate the total amount of money after 19 weeks.
In order to calculate this value we must replace w = 19 in the equaton
[tex]\begin{gathered} S=650+(19)(30) \\ S=1220 \end{gathered}[/tex]So, after 17 weeks there are $1220
I’m working on review questions to prepare for test and need help with this.
SOLUTION
Step1: write out the expression
[tex]\sqrt[]{125x}[/tex]we have to write this expression in its simplest form to identify the value of A and B
Step2: Identify the perfect square and write has a product
[tex]125=25\times5[/tex]Step3: replace the product above with 125 in the expression
[tex]\sqrt[]{125x}=\sqrt[]{25\times5\times x}[/tex]Step4: simplify the last expression by applying the rational rule below
[tex]\sqrt[]{a\times b}=\sqrt[]{a}\times\sqrt[]{b}[/tex]Hence we have
[tex]\sqrt[]{25\times5\times x}=\sqrt[]{25}\times\sqrt[]{5}\times\sqrt[]{x}[/tex][tex]\sqrt[]{25}\times\sqrt[]{5}\times\sqrt[]{x}=5\times\sqrt[]{5}\times\sqrt[]{x}=5\sqrt[]{5x}[/tex]Hence
[tex]\sqrt[]{125x}=5\sqrt[]{5x}[/tex]Therefore
A=5 and B=5x
20. Solve the equation, 3/5 * a = 1/4
We have to solve the equation:
[tex]\begin{gathered} \frac{3}{5}a=\frac{1}{4} \\ a=\frac{1}{4}\cdot\frac{5}{3} \\ a=\frac{5}{12} \end{gathered}[/tex]We just divide both sides by 5/3 and multiply. One side is clear a, and the other side results in 5/12.
Answer: a=5/12 (Option C)
ET 170 ivel -47 1) 0> Erol Name Kuta Software - Infinite Algebra 2 Solving Absolute Value Equations Solve each equation. 1) |3x| = 9 2) |–3r\ = 3 ond -3 Rasm
Question:
Solution:
Consider the following equation:
To solve the above equation, split the equation up into two separate equations:
Equation 1:
[tex]-2n\text{ + 6 = 6}[/tex]Equation 2:
[tex]-2n\text{ + 6 =- 6}[/tex]now, solve each of the equation:
For equation 1 we have that:
[tex]-2n\text{ = 0}[/tex]then n = 0
and for equation 2, we have that:
[tex]-2n\text{ = -12}[/tex]then n = 6.
Then, the correct answer is :
n = 6 and n = 0.
Hello! High school student in Calculus here. I need help solving the problem attached in the image. I'd like help FINDING THE DERIVATIVE using: - PRODUCT RULE- QUOTIENT RULEIf someone could help break down the steps and explain how to solve the problem using BOTH rule methods (because my teacher advised us to learn how to solve both ways), I would greatly appreciate it!
The function we have is:
[tex]h(\alpha)=\alpha^2\tan \alpha[/tex]Let's find the derivative using the product rule and the quotient rule.
-Using the product rule.
This is the better method for finding the derivative of this function because the function is already a product between two functions, which we will call f(alpha) and g(alpha):
[tex]\begin{gathered} f(\alpha)=\alpha^2 \\ g(\alpha)=\tan \alpha \end{gathered}[/tex]The product rule we will use is as follows:
[tex]h^{\prime}(\alpha)=f(\alpha)\times g^{\prime}(\alpha)+f^{\prime}(\alpha)\times g(\alpha)[/tex]Where the ' means derivative.
Let's first calculate the derivative of function f and g:
[tex]f(\alpha)=a^2\longrightarrow f^{\prime}(\alpha)=2\alpha[/tex]here we have used the following rule of derivation:
[tex]f(x)=x^{n^{}}\longrightarrow f^{\prime}(x)=nx^{n-1}[/tex]Now we find the derivative of g:
[tex]g(x)=\tan \alpha\longrightarrow g^{\prime}(\alpha)=\sec ^2\alpha[/tex]This is a derivate that we can find tables of derivation.
Now, we apply the product rule mentioned earlier:
[tex]h^{\prime}(\alpha)=f(\alpha)\times g^{\prime}(\alpha)+f^{\prime}(\alpha)\times g(\alpha)[/tex]And substitute all of the known values:
[tex]h^{\prime}(\alpha)=\alpha^2\sec ^2\alpha+2a\tan \alpha[/tex]That is our final result.
We can also calculate this using the quotient rule.
As we mentioned at the beginning, since there is a multiplication it is better to use the product rule to find the derivative, but we can also find it by using the quotient rule.
To use the quotient rule there need to be a division in the function, so we modify the original function:
[tex]h(\alpha)=\alpha^2\tan \alpha[/tex]And express it as a division as follows:
[tex]h(\alpha)=\frac{\alpha^2\sin\alpha}{\cos\alpha}[/tex]This is because
[tex]\tan \alpha=\frac{\sin \alpha}{\cos \alpha}[/tex]To use the quotient rule, we need to take or function:
[tex]h(\alpha)=\frac{\alpha^2\sin\alpha}{\cos\alpha}[/tex]And define one function f to be the numerator, and another function g to be the denominator:
[tex]f(\alpha)=\alpha^2\sin \alpha[/tex][tex]g(\alpha)=\cos \alpha[/tex]And the quotient rule is:
[tex]h^{\prime}(\alpha)=\frac{g(\alpha)f^{\prime}(\alpha)-f(\alpha)g^{\prime}(\alpha)}{g(\alpha)^2}[/tex]Before we find this, we need the derivatives of g and f:
[tex]g(\alpha)=\cos \alpha\longrightarrow g^{\prime}(\alpha)=-\sin \alpha[/tex]And for the derivative of f we apply the product rule because there is a multiplication:
[tex]f(\alpha)=a^2\sin \alpha\longrightarrow f^{\prime}(\alpha)=a^2(\sin \alpha)^{\prime}+(a^2)^{\prime}\sin \alpha[/tex]Solving the derivatives of this product rule:
[tex]f^{\prime}(\alpha)=a^2\cos \alpha+2a\sin \alpha[/tex]And now we are ready to use the quotient rule:
[tex]h^{\prime}(\alpha)=\frac{g(\alpha)f^{\prime}(\alpha)-f(\alpha)g^{\prime}(\alpha)}{g(\alpha)^2}[/tex]Substituting the known values:
[tex]h^{\prime}(\alpha)=\frac{\cos \alpha(a^2\cos \alpha+2a\sin \alpha)-a^2\sin \alpha(-\sin \alpha)}{\cos ^2\alpha}[/tex]Simplifying the expression:
[tex]h^{\prime}(\alpha)=\frac{a^2\cos^2\alpha+2a\sin\alpha\cos\alpha+a^2\sin^2\alpha}{\cos^2\alpha}[/tex]We factor the terms that contain a^2:
[tex]h^{\prime}(\alpha)=\frac{a^2(\cos ^2\alpha+\sin ^2\alpha)+2a\sin \alpha\cos \alpha}{\cos ^2\alpha}[/tex]Here, we use the following property of the cosine and the sine:
[tex]\cos ^2\alpha+\sin ^2\alpha=1[/tex]And we simplify again:
[tex]\begin{gathered} h^{\prime}(\alpha)=\frac{a^2+2a\sin \alpha\cos \alpha}{\cos ^2\alpha} \\ h^{\prime}(\alpha)=\frac{a^2}{\cos^2\alpha}+\frac{2a\sin \alpha\cos \alpha}{\cos ^2\alpha} \end{gathered}[/tex]We simplify further using:
[tex]\frac{1}{\cos^2\alpha}=\sec ^2\alpha[/tex][tex]h^{\prime}(\alpha)=a^2\sec ^2\alpha+\frac{2a\sin\alpha}{\cos\alpha}[/tex]which is equal to:
[tex]h^{\prime}(\alpha)=a^2\sec ^2\alpha+2a\tan \alpha[/tex]Answer:
[tex]h^{\prime}(\alpha)=a^2\sec ^2\alpha+2a\tan \alpha[/tex]solve by substitution You are planning a birthday party. You buy a total of 50 turkey burgers and veggie burgers for $90.00 . Yo pay $2.00 per Turkey burger and $1.50 per veggie burger. How many of each burger did you buy?*use verbal model to write a system of linear equations. Let "x" representthe number of turkey burgers and let "Y" represent the number of veggie burgers
Let,
t = Turkey burgers
v = veggie burgers
t + v = 50
$2.00 per Turkey burger
$1.50 per Veggie burger
2t + 1.5v = 90
Now we have a system of two equations with two unknowns
[tex]\begin{gathered} t+v=50 \\ 2t+1.5v=90 \end{gathered}[/tex]First equation
[tex]t=50-v[/tex][tex]\begin{gathered} 2(50-v)+1.5v=90 \\ 100-2v+1.5v=90 \\ -2v+1.5v=-100+90 \\ -0.5v=-10 \\ v=\frac{10}{0.5} \\ v=20 \end{gathered}[/tex]20 Veggie burgers
[tex]\begin{gathered} t=50-v \\ t=50-20 \\ t=30 \end{gathered}[/tex]30 Turkey burgers
Give each trig ratio as a fraction in simplest form. PQ is 48.
sin Q = 7/25
cos Q = 24/25
tan Q = 7/24
sin R = 24/25
cos R = 7/25
tan R = 24/7
Explanation:[tex]\begin{gathered} when\text{ Angle = Q} \\ \text{opposite = side opposite the angle= PR} \\ PR\text{ = 14} \\ \text{hypotenuse = 50} \\ \\ \sin \text{ Q = }\frac{opposite}{hypotenuse} \\ \sin \text{ Q = }\frac{14}{50} \\ \sin \text{ Q = 7/25} \end{gathered}[/tex][tex]\begin{gathered} \cos \text{ Q = }\frac{\text{adjacent}}{\text{hypotenuse}} \\ adjacent\text{ = PQ = ?} \\ \text{To get adjacent, we will apply pythagoras' theorem:} \\ \text{hypotenuse}^2=opposite^2+adjacent^2 \\ 50^2=14^2\text{ }+adjacent^2 \\ adjacent^2=50^2-14^2\text{ = 2500 - }196 \\ adjacent^2=\text{ 2304} \\ \text{adjacent = }\sqrt[]{2304}\text{ = 48} \\ \\ \cos \text{ Q = }\frac{48}{50} \\ \cos \text{ Q = 24/25} \end{gathered}[/tex][tex]\begin{gathered} \tan \text{ Q = }\frac{opposite}{adjacent} \\ \tan \text{ Q = }\frac{14}{48} \\ \tan \text{ Q = 7/24} \end{gathered}[/tex]when angle = R
opposite = side opposite the angle R = PQ
opposite = PQ = 48
adjacent = 14
hypotenuse = 50
[tex]\begin{gathered} \sin \text{ R = }\frac{opposite}{hypotenuse} \\ \sin \text{ R = }\frac{48}{50} \\ \sin \text{ R = 24/25} \end{gathered}[/tex][tex]\begin{gathered} \cos \text{ R = }\frac{\text{adjacent}}{\text{hypotenuse}} \\ \text{cos R = }\frac{14}{50} \\ \cos \text{ R = 7/25} \end{gathered}[/tex][tex]\begin{gathered} \tan \text{ R = }\frac{opposite}{hypotenuse} \\ \tan \text{ R = }\frac{48}{14} \\ \tan \text{ R = 24/7} \end{gathered}[/tex]I need help with this homework question please and thankyou
The formula for continuously compounded interest is
[tex]\begin{gathered} A=Pe^{rt} \\ \text{ Where }A\text{ is the Amount or future value} \\ P\text{ is the Principal, or initial value} \\ r\text{ is the interest rate, and} \\ t\text{ is the time} \end{gathered}[/tex]So, in this case, we have
[tex]\begin{gathered} A=\text{ \$}1,000 \\ P=\text{ \$}200 \\ r=4\text{\% }=\frac{4}{100}=0.04 \\ t=\text{ ?} \end{gathered}[/tex][tex]\begin{gathered} A=Pe^{rt} \\ \text{ Replace the know values} \\ \text{\$}1,000=\text{\$}200\cdot e^{0.04t} \\ \text{ Divide by \$200 from both sides of the equation} \\ \frac{\text{\$}1,000}{\text{\$}200}=\frac{\text{\$}200\cdot e^{0.04t}}{\text{\$}200} \\ 5=e^{0.04t} \\ \text{ Apply natural logarithm to both sides of the equation} \\ \ln (5)=\ln (e^{0.04t}) \\ \ln (5)=0.04t \\ \text{ Divide by 0.04 from both sides of the equation} \\ \frac{\ln(5)}{0.04}=\frac{0.04t}{0.04} \\ \boldsymbol{40.2\approx t} \end{gathered}[/tex]Therefore, it will take approximately 40 years for the account to reach $1,000.
In which direction does the left side of the graph of this function point?f(x) = 3x3 - x2 + 4x - 2Answer hereSUBMIT
The given function is expressed as
f(x) = 3x^3 - x^2 + 4x - 2
This is a cubic function which also means that it is an odd function.
The graph of the function is shown below
Write a rule to describe a transformation when a triangle hasbeen translated down 2 units and right 5 units.Using (x+10,y+10)
translated down 2 units and right 5 units.
The rule would be
(x + 5, y - 2)
Moving to the right is equal to adding 5 in the x-coordinate.
Translation down is equal to subtracting 2 in the y-coordinate.
A. Unfortunately, the precise data used by Eratosthenes was lost long ago. However, if Eratosthenes used a meter stick for his experiment today, then the stick’s shadow in Alexandria would be 127 mm long. Determine the angle 0 that the sunrays made with the meter stick. Remember that a meter stick is 1000 millimetres long.B. Assuming that the sun’s rays are essentially parallel, determine the central angle of the circle if the angle passes through Alexandria and Syene. How did you find your answer?
A
Answer:
Explanation:
The right triangle formed is shown below.
To find θ, we would apply the tangent trigonometric ratio which is expressed as
tan θ = opposite side /adjacent side
From the triangle,
opposite side = 127
adjacent side = 1000
tanθ = 127/1000 = 0.127
taking the tan inverse of 0.127
θ = tan^1(0.127)
θ = 7 degrees
The perimeter of a rectangle measuring (2x + 2)cm by (3x -3) cm is 58cm. Calculate its area. 13. The perimeter
Given:
The length of the given rectangle is l =(2x+2) cm.
The width of the given rectangle is w =(3x-3) cm.
The perimeter of the given rectangle is P =58cm.
Required:
We need to find the area of the rectangle.
Explanation:
Consider the perimeter of the rectangle formula.
[tex]P=2(l+w)[/tex]Substitute l=2x+2, w =3x-3 and P=58 in the formula.
[tex]58=2(2x+2+3x-3)[/tex][tex]58=2(5x-1)[/tex][tex]58=10x-2[/tex]Add 2 to both sides of the equation.
[tex]58+2=10x-2+2[/tex][tex]60=10x[/tex]Divide both sides by 10.
[tex]\frac{60}{10}=\frac{10x}{10}[/tex][tex]x=6[/tex]Substitute x =6 in the equations l=2x+2.
[tex]l=2x+2=2(6)+2=12+2=14cm[/tex][tex]Substitute\text{ }x=6\text{ }in\text{ }the\text{ }equations\text{ }w=3x-3.[/tex][tex]w=3x-3=3(6)-3=18-3=15cm[/tex]The area of the rectangle is
[tex]A=lw[/tex]Substitute l=14cm and w =15cm in the formula.
[tex]A=14\times15[/tex][tex]A=210cm^2[/tex]Final answer:
The area of the given rectangle is 210 square cm.
The two rectangles below have the same height-to-width ratio. What is thevalue of w? (If necessary, round your answer to two decimal places.)21W=OA. 11.33OB. 17OC. 20OD. 4.3234
Solution
- Since the height-to-width ratio of both rectangles are the same, we can solve the question as follows:
[tex]\begin{gathered} \frac{H}{W}=\frac{h}{w} \\ where, \\ H=height\text{ of big rectangle} \\ W=width\text{ of big rectangle} \\ h=height\text{ of small rectangle} \\ w=width\text{ of small rectangle} \\ \\ \text{ Thus, we have:} \\ \frac{21}{34}=\frac{7}{w} \\ \\ \text{ Make w the subject of the formula} \\ w=\frac{34\times7}{21} \\ \\ w=11.333\bar{3} \end{gathered}[/tex]Final Answer
OPTION A
help me graduate please the graph of each function is shown . write the function in factored format. do not include complex numbers
Given the function of the graph:
[tex]g(x)=2x^4+x^3-47x^2-25x-75[/tex]Let's factor the given function.
Regroup the terms:
[tex]g(x)=x^3-25x+2x^4-47x^2-75[/tex][tex]\begin{gathered} \text{Factor x out of x}^3-25x\colon \\ \\ g(x)=x(x^2-25)+2x^4-47x^2-75 \end{gathered}[/tex][tex]g(x)=x(x^2-5^2)+2x^4-47x^2-75[/tex][tex]g(x)=x(x+5)(x-5)+2x^4-47x^2-75[/tex][tex]\begin{gathered} \text{ Rewrite x}^4as(x^2)^2\colon \\ \\ g(x)=x(x+5)(x-5)+2(x^2)^2-47x^2-75 \end{gathered}[/tex][tex]\begin{gathered} Letu=x^2 \\ \\ g(x)=x(x+5)(x-5)+2u^2-47u^{}-75 \end{gathered}[/tex][tex]\begin{gathered} Factor\text{ by grouping:} \\ g(x)=x(x+5)(x-5)+(2u+3)(u-25) \end{gathered}[/tex][tex]\begin{gathered} \text{ Repalce u with x}^2\colon \\ g(x)=x(x+5)(x-5)+(2x^2+3)(x^2-25) \end{gathered}[/tex][tex]g(x)=x(x+5)(x-5)+(2x^2+3)(x^2-5^2)[/tex][tex]g(x)=x(x+5)(x-5)+(2x^2+3)(x+5)(x-5)[/tex][tex]\begin{gathered} \text{Factor out (x+5)(x-5)} \\ \\ g(x)=(x+5)(x-5)(x+2x^2+3) \\ \\ g(x)=(x+5)(x-5)(2x^2+x+3) \end{gathered}[/tex]ANSWER:
[tex]g(x)=(x+5)(x-5)(2x^2+x+3)[/tex]Represent the following sentence as an algebraic expression, where "anumber" is the letter x.The difference of a number and 8.
Given,
The difference of a number and 8
Let the number be x
The result of subtracting one number from another is called the "Difference"
The difference of a number and 8 can be expressed below as
[tex]x-8[/tex]Hence, the algebraic expression of the difference of a number (x) and 8 is x - 8
in circle C, BC = 6 and angle ACB = 120 degrees. what is the area of the shaded sector?
Area of sector = θ/360 × πr²
θ = m∠ACB = 120°
radius = r = 6
[tex]\text{Area of sector = }\frac{120}{360}\times\pi\times6^2[/tex][tex]\begin{gathered} \text{Area = }\frac{1}{3}\times\pi\times36 \\ Area\text{ = }\pi\times\frac{36}{3}\text{ } \end{gathered}[/tex][tex]\begin{gathered} Area\text{= }\pi\times12 \\ \text{Area of sector = 12}\pi\text{ (option C)} \end{gathered}[/tex]If g is a linear function and g(2)=7 and g(-2)=-1, find g(-5)
Since the function is linear, it can be written in slope intercept form which is expressed as
y = mx + c
where
m = slope
c = y intercept
The formula for calculating slope is expressed as
m = (y2 - y1)/(x2 - x1)
From the information given,
g(2) = 7
This means that if x2 = 2, y2 = 7
Also,
g(- 2) = - 1
This means that if x1 = - 2, y1 = - 1
By substituting these values into the formula for calculating slope, we have
m = (7 - - 1)/(2 - - 2) = (7 + 1)/(2 + 2) = 8/4 = 2
We would find the y intercept, c by substituting x = 2, y = 7 and m = 2 into the slope intercept equation. We have
7 = 2 * 2 + c
7 = 4 + c
c = 7 - 4 = 3
By substituting m = 2 and c = 3 into the slope intercept equation, the linear function is
y = 2x + 3
Writing it as a function in terms of g, it is
g(x) = 2x + 3
To find g(- 5), we would substitue x = - 5 into g(x) = 2x + 3, we have
g(- 5) = 2(- 5) + 3 = - 10 + 3
g(- 5) = - 7
Harrison saved $38.97 from his first paycheck and $65.04 from his second paycheck. How much did he save from the paychecks?A.$26.07B.$94.01C.$94.91D.$104.01
Answer:
D.$104.01
Explanation:
Here is what we are told
Harrison got the following amount from his paychecks.
Paycheck 1: $38. 97
Paycheck 2: $65.04
Therefore total amount Harrison saved on his paychecks was
$38. 97 + $65.04 = $104.01
Now looking at the answer choices we see that choice D gives the correct answer.
Therefore choice D is the correct answer.
Answer:
it’s D sir =)
Step-by-step explanation:
Directions: for questions 1 through 4, find the diagonal length of each solid figure.
Here, we want to find the diagonal of the given solid
To do this, we need the appropriate triangle
Firstly, we need the diagonal of the base
To get this, we use Pythagoras' theorem for the base
The other measures are 6 mm and 8 mm
According ro Pythagoras' ; the square of the hypotenuse equals the sum of the squares of the two other sides
Let us have the diagonal as l
Mathematically;
[tex]\begin{gathered} l^2=6^2+8^2 \\ l^2\text{ = 36 + 64} \\ l^2\text{ =100} \\ l\text{ = }\sqrt[]{100} \\ l\text{ = 10 mm} \end{gathered}[/tex]Now, to get the diagonal, we use the triangle with height 5 mm and the base being the hypotenuse we calculated above
Thus, we calculate this using the Pytthagoras' theorem as follows;
[tex]\begin{gathered} d^2=5^2+10^2 \\ d^2\text{ = 25 + 100} \\ d^2\text{ = 125} \\ d\text{ = }\sqrt[]{125} \\ d\text{ = }11.2\text{ mm} \end{gathered}[/tex]A cone has a volume of 367 cm and height h. Complete this table for volume of cylinders with the same radius but different heights. height (cm) h volume (cm) 367 2h 5h h 2 Write your answer in terms of 77, just like the first example 367. When typing your answer, use the word "pi" to represent the symbol 7. For example, 36 would be typed as 36pi
If f(x) = x^2 + 3x, find f(-3). *
Answer:
The value of f(-3) is;
[tex]f(-3)=0[/tex]Explanation:
Given that;
[tex]f(x)=x^2+3x[/tex]To get f(-3), we will substitute -3 for x in the given function f(x);
[tex]\begin{gathered} f(x)=x^2+3x \\ f(-3)=(-3)^2+3(-3) \\ f(-3)=9^{}-9 \\ f\mleft(-3\mright)=0 \end{gathered}[/tex]Therefore, the value of f(-3) is;
[tex]f(-3)=0[/tex]Can you pls help me with this question thank you
The given expression is:
[tex]16+\frac{r^2}{s^2}[/tex]When r=6 and s=2, we need to replace those values into the expression and solve:
[tex]16+\frac{6^2}{2^2}=16+\frac{36}{4}=16+9=25[/tex]The answer is a. 25
What is 35/12 written as a Mixed Number?
Answer:
2 11/12
Explanation:
To write 35/12 as a mixed number we need to divide 35 by 12.
When we divide 35 by 12, we get 2 as a quotient and 11 as a remainder.
Additionally, 35 is the dividend, and 12 is the divisor:
Now, the mixed number can be written as:
It means that the mixed number is:
[tex]\text{quotient}\frac{\text{ remainder}}{\text{divisor}}=2\frac{11}{12}[/tex]Therefore, 35/12 written as a mixed number is 2 11/12
Enter the range of values for x using the picture shown
We are asked to find a range for x. Let's start by finding the maximum value. We know, by definition, that a side opposite a larger angle has a longer length. For example, if you think about a right triangle, the hypotenuse (the longest side) is always opposite of the right angle (the largest angle). We are given two side lengths: 23 is opposite of 42 degrees, and 21 is opposite of 3x + 15. Because 42 is opposite the longer side, 42 degrees is a larger angle than 3x + 15. We can set up the inequality and solve for x:
[tex]\begin{gathered} 3x+15<42 \\ 3x<27 \\ x<9 \end{gathered}[/tex]Now, let's look at the minimum value. We know that the angle definitely has to be larger than 0. So, we can set up that inequality and solve for x:
[tex]\begin{gathered} 3x+15>0 \\ 3x>-15 \\ x>-5 \end{gathered}[/tex]Now, we have our final range: -5 < x < 9
A polygon has a perimeter of 18 units. It is dilated with a scale factor of . What is theperimeter of its image?
The new perimeter is as follows:
[tex]P=18\cdot\frac{3}{2}=9\cdot3=27[/tex]We multiply the former perimeter by the scale factor. Thus, it should be 27 units.
In the similar triangles below, what is the length of AB?
Given:
There are given two triangles, ABC and DEF.
Where,
[tex]\begin{gathered} AC=6cm \\ BC=4cm \\ DE=15cm \\ DF=18cm \end{gathered}[/tex]Explanation:
To find the value of two congruent triangles, we need to use the ratio properties:
So,
From the given congruent triangle:
[tex]\frac{AB}{DE}=\frac{AC}{DF}[/tex]Then,
Put the all values into the above ratio expression:
So,
[tex]\begin{gathered} \begin{equation*} \frac{AB}{DE}=\frac{AC}{DF} \end{equation*} \\ \frac{AB}{15}=\frac{6}{18} \\ \frac{AB}{15}=\frac{1}{3} \\ 3AB=15 \\ AB=\frac{15}{3} \\ AB=5 \end{gathered}[/tex]Final answer:
Hence, the correct option is D.