Explanations:
Price of a cup of Ice cream = $2
Price of one topping = $0.75
Since Justin wants four toppings on the cup of ice cream that he buys:
Price of four toppings = $0.75 * 4
Price of four toppings = $3.00
The total cost of Ice cream bought by Justin = Price of a cup of Ice cream + Price of four toppings
The total cost of Ice cream bought by Justin = $2 + $3.00
The total cost of Ice cream bought by Justin = $5.00
Solve equation: 2x^2-50=0
Given:
[tex]2x^2-50=0[/tex]To solve for x:
[tex]\begin{gathered} 2x^2-50=0 \\ 2x^2=50 \\ x^2=25 \\ x=\pm5 \end{gathered}[/tex]Hence, the solution is x=-5, 5.
Answer:
[tex]x=[/tex] ± [tex]5[/tex]
Step-by-step explanation:
[tex]2x^{2} -50=0[/tex]
[tex]2x^{2} =50[/tex]
[tex]x^{2} =\frac{50}{2}[/tex]
[tex]x^{2} =25[/tex]
[tex]x=\sqrt{25}[/tex]
[tex]x=[/tex] ± [tex]5[/tex]
There are 2 solutions for x: 5 positive and 5 negative. If you square them they result in 25
Hope this helps
O GRAPHS AND FUNCTIONSTransforming the graph of a function by shrinking or stretching
A.
The graph of y= f(0.5x) invoves shrinking the graph by a factor of 0.5
This is shown below
how would I find slope
the formula to find a slope between two points is
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]if the equation of a line is given
y = mx +c
then m is the slope of the line.
Why must the first and last term of a perfect square trinomial be both positive?
Given:
The objective is to provide the reason for first and last term of a perfect square trinomial be both positive
Explanation:
Consider the following equations,
[tex]\begin{gathered} (a+b)^2=(+a)^2+2(a)(b)+(+b)^2\text{ . . . . . . . (1)} \\ (a-b)^2=(+a)^2+2(a)(b)+(-b)^2\text{ . . . . . . . (2)} \\ (-a+b)^2=(-a)^2+2(a)(b)+(+b)^2\text{ . . . . . . . (3)} \end{gathered}[/tex]From equations (1), (2) and (3), the first and last term are always perfect squares. So, even when those numbers are negative integer, square of the number will get converted into poitive integer.
Hence, the first and the last term of a perfect square trinomial must be positive.
Use the given scale factor and the side lengths of the scale drawing to determine the side lengths of the real object. Scale factor: 2:1 5 in 10 in 6 C с 13 in Scale drawing Object O A. Side a is 11 inches long, side bis 8 inches long, and side cis 3 inches long. O B. Side a is 26 inches long, side bis 20 inches long, and side cis 10 inches long. O C. Side a is 2.5 inches long, side bis 5 inches long, and side cis 6.5 inches long. D. Side a is 15 inches long, side bis 12 inches long, and side 7
If the scale factor is 2: 1
a = 5/2 = 2.5 inches
b = 10/2 = 5 inches
c = 13/5 = 6.5 inches
Please help me on number one It’s all one question
Given:
[tex]f(x)=2x^2+12x+10[/tex]a) Standard form of the function is,
[tex]\begin{gathered} f(x)=2x^2+12x+10 \\ f(x)=2(x^2+6x+5+9-9) \\ f(x)=2(x^2+6x+9-4) \\ f(x)=2(x+3)^2-8 \end{gathered}[/tex]Standard form is,
[tex]f(x)=2(x+3)^2-8[/tex]b) The vertex of the given function
The vertex of the parabola having form,
[tex]\begin{gathered} y=ax^2+bx+c \\ \text{Vertex}=\frac{-b}{2a} \\ f(x)=2x^2+12x+10 \\ \text{Vertex}=\frac{-b}{2a}=\frac{-12}{2(2)}=-\frac{12}{4}=-3 \\ \text{Put x=-3 in }f(x)=2x^2+12x+10 \\ f(x)=2(-3)^2+12(-3)+10=18-36+10=-8 \end{gathered}[/tex]Vertex is ( -3,-8)
c) x and y-intercept is,
[tex]\begin{gathered} Set\text{ y=0 that means f(x)=0} \\ f(x)=2x^2+12x+10 \\ 2x^2+12x+10=0 \\ 2(x^2+6x+5)=0 \\ x^2+6x+5=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},a=1,b=6,c=5 \\ x=\frac{-12\pm\sqrt{12^2-4\cdot\:2\cdot\:10}}{2\cdot\:2} \\ x=\frac{-12\pm\: 8}{4} \\ x=-1,x=-5 \end{gathered}[/tex]x- intercepts are (-1,0) and (-5,0).
For y-intercept , set x=0
[tex]\begin{gathered} f(x)=2x^2+12x+10 \\ y=2x^2+12x+10 \\ y=2(0)+12(0)+10 \\ y=10 \end{gathered}[/tex]y-intercept is (0,10)
d) the graph of the function is,
e) The domain and range of the function is,
[tex]\begin{gathered} \text{For range of the function f(x)=ax}^2+bx+c\text{ with vertex (x,y)} \\ \text{If a}<0\text{ range is }f(x)\leq y \\ \text{If a>0, range is f(x)}\ge\text{y} \end{gathered}[/tex]For the given function,
[tex]\begin{gathered} f(x)=2x^2+12x+10\text{ with vertex (-3,-8)} \\ a=2>0,\text{ range is f(x)}\ge\text{-8} \\ \text{Domain is -}\inftyTherefore,[tex]\begin{gathered} \text{Domain of f is (-}\infty,\infty) \\ \text{Range of f is \lbrack-8,}\infty) \end{gathered}[/tex]I need help with question 8 i just need the answer I just want to see if I’m right
Answer:
No
Explanation:
The function is one-to-one if when we draw a horizontal line in the graph, the line crosses the function once. In this case, we get:
Since the line crosses the graph twice, the given function is not one-to-one.
So, the answer is No.
Given ΔQRS≅ΔTUV, QS=3v+2andTV=7v−6, find the length of QS and TV.
Since the triangles ΔQRS and ΔTUV, their corresponding sides are also congruent, this means that
[tex]\begin{gathered} QR\cong TU \\ RS\cong UV \\ QS\cong TV \end{gathered}[/tex]Since QS and TV are congruent, this means that they have the same measures, therefore
[tex]\begin{gathered} QS=TV \\ 3v+2=7v-6 \end{gathered}[/tex]Solving this equation for v, we have
[tex]\begin{gathered} 3v+2=7v-6 \\ 3v-7v+2=-6 \\ -4v=-6-2 \\ 4v=8 \\ v=2 \end{gathered}[/tex]Using this value for v in our expression for the sides, we have
[tex]\begin{gathered} QS=3\cdot2+2=8 \\ TV=7\cdot2-6=8 \end{gathered}[/tex]Both sides are equal to 8.
How do you know when to factor a negative square root number and when to just find it’s imaginary number
sqrt(-44)
This has a perfect square inside the square root
4 is a perfect square
sqrt(-1 * 4 * 11)
sqrt(-1) sqrt(4) sqrt(11)
We know the square root of 4 is 2
2 sqrt(11) sqrt(-1)
2 sqrt(11) i
sqrt(-30)
sqrt( 6*5* -1)
There is no perfect square for 30
Another example with a perfect square inside would be
sqrt(-54)
sqrt( -1 * 27*2)
sqrt( -1 * 9 *6)
9 is a perfect square
sqrt( -1) * sqrt(9) sqrt(6)
3 i sqrt(6)
Try to break down larger numbers and see if there is a perfect square that is a factor
Write an equivalent unit rate to eating 4 hot dogs in 1/3 of a minute
To write an equivalent unit rate, we have to divide the number of hot dogs by the number of minutes, as follows:
[tex]\frac{4\text{ hot dogs}}{\frac{1}{3}\text{ minutes}}=3\cdot4\frac{\text{ hot dogs}}{\text{ minute}}=12\frac{\text{ hot dogs}}{\text{ minute}}[/tex]the perimeter of a rectangle is 54 inches the length is three more than twice the width what are the dimensions of the rectangle
Perimeter of a rectangle = (2x length) + ( 2 x width)
if the length is 3 more than twice the width then
length = 3 +( 2x width)
substituting for the perimeter which is given as 54 inches and the length derived as above in to the equation for the perimeter. we have
54 = 2(3+(2 x w)) + 2w
54 = 6 +4w +2w
54 =6 +6w
6w = 48
6w/6 = 48/6
w = 8 inches
Substituting for w in the equation for length gives us
length = 3 + (2 x8)
L= 3+16
L = 18 inches
The dimesnsions of the rectangle are therefore
W = 8 inches
and
L = 18 inches
A cyclist rode for 3.5 hours and completed a distance of 60.9 miles. If she kept the same average speedfor each hour, how far did she ride in 1 hour?Part AEstimate the quotient. Include an equation to show your work. Explain your thinking.Math symbols+.808(.)O=<>+$?Part BFind the exact distance she rode in 1 hour by completing the division Enter the answer in the box10 of 14 Answered
We know that the cyclist rode for 3.5 hours and completed a distance of 60.9 miles.
We can estimate the average speed as the quotient between the distance and the time:
[tex]\bar{v}=\frac{d}{t}=\frac{60.9\text{ miles}}{3.5\text{ hours}}[/tex]We have now to estimate the distance she will ride in one hour.
We can estimate the average speed as:
[tex]\frac{60.9}{3.5}\approx\frac{60\cdot2}{3.5\cdot2}=\frac{120}{7}\approx17[/tex]She will ride 17 miles in an hour.
If we solve the quotient 60.9 / 3.5 with a calculator, we get 17.4 miles per hour.
Answer:
a) We can estimate that she will ride 17 miles in an hour.
b) 17.4 miles per hour.
Answer:
17.4
Step-by-step explanation:
A line connects points A and B as shown below. If a third point, C, is plotted six units to the right of point B, what will be the area of the resulting right triangle?O 48O 24O 30O 18
ANSWER
EXPLANATION
We are given the two points A and B connected by a line.
A third point C is plotted 6 units to the right of B.
So, we have:
Graph -9x + 5y = 45. Intro y Unit Test skills 9 8+ 7+ 6 CO 5 4 3 No 1 1 х -9-8-7 -6 -5 -4 -3 -2 1 2 3 4 5 6 7 8 9 -2 -3 -4 -5 -6 -7 -8 -9
Here, we want to graph the given line
Firstly, we write the equation in the standard form
The standard form is;
[tex]y\text{ = mx + b}[/tex]where m represents the slope and b is the y-intercept
We have this as;
[tex]\begin{gathered} 5y\text{ = 9x + 45} \\ \text{divide through by 5} \\ y\text{ = }\frac{9}{5}x\text{ + }\frac{45}{5} \\ \\ y\text{ = 1.8x + 9} \end{gathered}[/tex]we have the y-intercept, this is 9, so we mark the point (0,9)
What is left is to get the possible x-intercept
As the value of y at this point is zero
We have it that;
[tex]\begin{gathered} 0\text{ = 1.8x + 9} \\ 1.8x\text{ = -9} \\ x\text{ = }\frac{-9}{1.8} \\ x\text{ = -5} \end{gathered}[/tex]The x-intercept is -5, which in the coordinate form is (-5,0)
So, by joining the points (-5,0) and (0,9), we have the graph of the line
A plot is shown below;
write an equation in point slope form of the line that passes through the given points or with the given characteristics
The equation of a line in point-slope form is
[tex](y-y_1)=m(x-x_1)[/tex]At m = -2 and (-3 , 5) we have,
[tex]\begin{gathered} (y-5)=-2(x--3) \\ (y-5)=-2(x+3) \end{gathered}[/tex]That is the equation in point-slope form.
Identify the independent and dependent variable of the following graph. Indicate whether the graph rises, faconstant.wAnnual gross income in relation to number of years.Annual Income
Given data:
The given graph.
The annual can can be written as,
[tex]\text{Annual income =f(time)}[/tex]The independent variable is time, the dependent variable is Annual income.The annual income graph rises steadily then drop off sharply before begining to rise again.
Thus, the option (d) is correct.
The total cost (c) in dollars of renting a car for d days is given by the equationc =50 + 10n. If the total cost was $130, for how many days was the car rented?A. 8B. 13C. 3D. 18
Given:
The equation of total cost is, c = 50+10n.
The total cost is, c = $130.
The objective is to find the number of days for renting the car.
Substitute the value of c in the given equation of total cost.
[tex]\begin{gathered} 130=50+10n \\ 130-50=10n \\ 80=10n \\ n=\frac{80}{10} \\ n=8 \end{gathered}[/tex]Thus, the number of days for renting the car is 8.
Hence, option (A) is the correct answer.
simplify 3x squared minus 6x plus 12 minus 6x plus 10
Given:
[tex]3x^2-6x+12-6x+10[/tex]Let's simplify:
Collect like terms and evaluate
[tex]3x^2-6x-6x+12+10[/tex][tex]3x^2-12x+22[/tex]ANSWER:
[tex]3x^2-12x+22[/tex]What are the first 5 multiples of 5 what are the first 5multiples of 2 what is the least common multiple(LCM) of 5 and 2
The first 5 multiples of 5 is
Multiples of 5:
5, 10, 15, 20, 25
The first 5 multiples of 2 is
Multiples of 2:
2, 4, 6, 8, 10
Therefore, the LCM of 5 and 2 is 10.
The---------------- is the annual amount you pay just to own a credit card.
You need to pay a yearly fee to own a credit card.
Moreover, you need to pay interest on the amount you spend.
This is how credit card works.
Also, the bank checks your credit before issuing a credit card.
The yearly fee, the annual fee, is charged at the end of every fiscal year.
Find the perimeter of the composite figureA) 18 ftB) 14 ftC) 16 ft D) 17 ft
The perimeter of a figure is the sum of all of its sides.
In the given figure we can observe two missing sides, so we need to find its lengths as follows:
So:
[tex]\begin{gathered} 3ft+y=4ft \\ y=4ft-3ft \\ y=1ft \\ And \\ x+2ft=5ft \\ x=5ft-2ft \\ x=3ft \end{gathered}[/tex]Thus, the perimeter is:
[tex]\begin{gathered} P=3ft+5ft+4ft+2ft+y+x \\ P=3ft+5ft+4ft+2ft+1ft+3ft \\ P=18ft \end{gathered}[/tex]The answer is A) 18 ft.
If a sample mean is 96, which of the following is most likely the range ofpossible values that best describes an estimate for the population mean?A. (82, 114)B. (76,108)O C. (78, 110)D. (80, 112)
Ok! I understand.
So, we have to find the average of each option between these points.
Let's start with A.
Average will be: (82+114)/2 which is equal to 98.
98 is not equal to 96, so that's not the answer.
Let's see B.
Average will be (76+108)/2 which is equal to 92.
92 is not equal to 96, so that's not the answer.
Let's see C.
this is a multiple choice question no need for it to be long
In order to determine which from the given equation represents the given data, first consider that while x increases y decreases. It means that the slope of the line is positive.
Furthermore, calculate the slope by using the following formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]where (x1,y1) and (x2,y2) are two points of the table. You can select any pair of points, for example:
(x1,y1) = (0,2)
(x2,y2) = (2,-4)
replace the previous values into the formula for m:
[tex]m=\frac{-4-2_{}}{2-0}=-\frac{6}{2}=-3[/tex]Hence, the slope of the line is -3.
Hence, you can conclude that the equation of the line that best respresents the given data is:
y = -3x + 2
help me find the slope plss
Find - 34 + 21 - 39 - ( - 3 )
-49
Explanation:[tex]-34\text{ + 21 - 39 -(-3)}[/tex]Expanding the parenthesis:
[tex]\begin{gathered} mu\text{ltiplication of asme sign gives a positive number:} \\ -34\text{ + 21 - 39 + 3} \end{gathered}[/tex]simplify:
[tex]\begin{gathered} =-34\text{ - 39 + 21 + 3} \\ -34\text{ - 39 = - 7}3 \\ 21\text{ + 3 = 24} \end{gathered}[/tex][tex]\begin{gathered} -39\text{ - 39 + 21 + 3 = -73 + 24}| \\ =\text{ }-49 \end{gathered}[/tex]This is one practice problem that I need help on It asks to solve logarithmic equations, and It includes a question within it. Please note that this is a lengthy problem. & I have enlarged the equations for better view.
The Solution:
We are required to solve each of the following logarithmic equations:
a.
[tex]\begin{gathered} 1.\text{ }\log _464=m \\ \text{cross multiplying, we get} \\ 64=4^m \\ 4^3=4^m \\ m=3 \end{gathered}[/tex][tex]\begin{gathered} 2.\text{ }\log _8n=3 \\ \text{Cross multiplying, we get} \\ n=8^3=512 \end{gathered}[/tex][tex]\begin{gathered} 3.\text{ }\log _p4096=3 \\ \text{Cross multiplying, we get} \\ 4096=p^3 \\ \text{ Equalising the base in both sides, we get} \\ 16^3=p^3 \\ 16=p \\ p=16 \end{gathered}[/tex][tex]\begin{gathered} 4.\text{ }\log _{32}q=3 \\ \text{cross multiplying, we get} \\ q=32^3=32\times32\times32=32768 \end{gathered}[/tex]b.
Given the equation below:
[tex]\log _{4^x}2^a=3[/tex]We are required to express a in terms of x.
[tex]\begin{gathered} \log _{4^x}2^a=3 \\ \text{Cross multiplying, we get} \\ 2^a=(4^x)^3 \\ \text{Equalizing the base, we get} \\ 2^a=(2^{2x})^3 \\ 2^a=2^{6x} \\ a=6x \end{gathered}[/tex]Module 1: Ratios and unit rates: Q... elect three ratios that are equivalent 16 : 12. hoose 3 answers: 8:6
16:12 is equivalent to 8:6 and is equivalent to 4:3
The three lines represent the number of hot dogs eaten by three contestants at a hot dog eating competition. Which contestant was eating at the slowest rate?A. Contestant J B. Contestant KC. Contestant L
B. Contestant K.
By looking at the graph we can see that the line K has the lowest slope ( change on y/ change on x)
List the domain and Range, and determine if it's a funtion or not
The graph is a function
Explanation:The domain are the x values of a function. They are the input of the function
From the graph, following the x axis:
The arrow on the line indicates it doesn't end.
As a result, it moves from negative infinity to positive infinity
[tex]Domain\colon(-\infty,\text{ }\infty)[/tex]Range are the y values of the function. They are the output of a function
From the graph, following the y axis:
the line starts from y = -3 towards the positive side of y axis.
Since there is an arrow on the line pointing upwards towards the y axis, it indicates positive infinity.
[tex]\text{Range = (-3, }\infty)[/tex]For it to be a function, the input will have only one output:
Each of the x values (input) gives only one y value (output)
The graph is a function
Find the derivative of y=3x/(x^2 + 1)
To find the derivative of
[tex]y=\frac{3x}{x^2+1},[/tex]we will use the division rule for derivatives.
The division rule states that:
[tex](\frac{f(x)}{g(x)})^{\prime}=\frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{g(x)^2}.[/tex]Therefore, the derivative of the given quotient is:
[tex]y^{\prime}=\frac{(3x)^{\prime}(x^2+1)-(x^2+1)^{\prime}(3x)}{(x^2+1)^2}.[/tex]Simplifying the above result we get:
[tex]\begin{gathered} y^{\prime}=\frac{3(x^2+1)-(2x\cdot3x)}{(x^2+1)^2}=\frac{3(x^2+1)-6x^2}{(x^2+1)^2}=\frac{3(x^2+1-2x^2)}{(x^2+1)^2} \\ =\frac{3(1-x^2)}{(x^2+1)^2}. \end{gathered}[/tex]Answer:
[tex]y^{\prime}=\frac{3(1-x^2)}{(x^2+1)^2}.[/tex]