Answer:
The missing side is [tex]B = 6.0\ cm[/tex]
The missing angles are [tex]\alpha = 56.2[/tex] and [tex]\theta = 93.8[/tex]
Step-by-step explanation:
Given
[tex]A = 10\ cm[/tex]
[tex]C = 12\ cm[/tex]
[tex]\beta = 30[/tex]
The implication of this question is to solve for the missing side and the two missing angles
Represent
Angle A with [tex]\alpha[/tex]
Angle B with [tex]\beta[/tex]
Angle C with [tex]\theta[/tex]
Calculating B
This will be calculated using cosine formula as thus;
[tex]B^2 = A^2 + C^2 - 2ACCos\beta[/tex]
Substitute values for A, C and [tex]\beta[/tex]
[tex]B^2 = 10^2 + 12^2 - 2 * 10 * 12 * Cos30[/tex]
[tex]B^2 = 100 + 144 - 240 * 0.8660[/tex]
[tex]B^2 = 100 + 144 - 207.8[/tex]
[tex]B^2 = 36.2[/tex]
Take Square root of both sides
[tex]B = \sqrt{36.2}[/tex]
[tex]B = 6.0[/tex] (Approximated)
Calculating [tex]\alpha[/tex]
This will be calculated using cosine formula as thus;
[tex]A^2 = B^2 + C^2 - 2BCCos\alpha[/tex]
Substitute values for A, B and C
[tex]A^2 = B^2 + C^2 - 2BCCos\alpha[/tex]
[tex]10^2 = 6^2 + 12^2 - 2 * 6 * 12 * Cos\alpha[/tex]
[tex]100 = 36 + 144 - 144Cos\alpha[/tex]
Collect Like Terms
[tex]100 - 36 - 144 = -144Cos\alpha[/tex]
[tex]-80 = -144Cos\alpha[/tex]
Divide both sides by -144
[tex]\frac{-80}{-144} = Cos\alpha[/tex]
[tex]0.5556 = Cos\alpha[/tex]
[tex]\alpha = cos^{-1}(0.5556)[/tex]
[tex]\alpha = 56.2[/tex] (Approximated)
Calculating [tex]\theta[/tex]
This will be calculated using cosine formula as thus;
[tex]C^2 = B^2 + A^2 - 2BACos\theta[/tex]
Substitute values for A, B and C
[tex]12^2 = 6^2 + 10^2 - 2 * 6 * 10Cos\theta[/tex]
[tex]144 = 36 + 100 - 120Cos\theta[/tex]
Collect Like Terms
[tex]144 - 36 - 100 = -120Cos\theta[/tex]
[tex]8 = -120Cos\theta[/tex]
Divide both sides by -120
[tex]\frac{8}{-120} = Cos\theta[/tex]
[tex]-0.0667= Cos\theta[/tex]
[tex]\theta = cos^{-1}(-0.0667)[/tex]
[tex]\theta = 93.8[/tex] (Approximated)
Question 6 of 10
Which equation matches the graph of the greatest integer function given
below?
Use the minimum and maximum data entries and the number of classes to find the class width, the lower class limits, and the upper class limits. min = 14, max = 121, 8 classes
Answer:
The class width is [tex]C_w \approx 13[/tex]
Step-by-step explanation:
From the question we are told that
The upper class limits is [tex]max = 121[/tex]
The lower class limits is [tex]min = 14[/tex]
The number of classes is [tex]n = 8 \ classes[/tex]
The class width is mathematically represented as
[tex]C_w = \frac{max - min}{n }[/tex]
substituting values
[tex]C_w = \frac{121 - 14}{8 }[/tex]
[tex]C_w = 13.38[/tex]
[tex]C_w \approx 13[/tex]
Since
A rectangle has length 4 inches and width 2 inches. If the length and width of the rectangle are
reduced by 50 percent, by what percent will the area of the rectangle be reduced?
40 percent
50 percent
60 percent
75 percent
Answer:
75%
Step-by-step explanation:
First we can solve the area of the rectangle originally the answer is:
4 × 2 = 8
Then we decrease both measurements by 50% to get the dimensions 1 and 2. The new area will be 1 × 2 which is 2.
2 is 25% of 8 which means that the area of the rectangle has been reduced by 75%.
The half-life of a radioactive isotope is the time it takes for a quantity of the Isotope to be reduced to half its initial mass. Starting with 210 grams of a
radioactive isotope, how much will be left after 6 half-lives?
Round your answer to the nearest gram
Answer:
3 grams will be left after 6 half-lives
Step-by-step explanation:
Half-live:
Time it takes for the substance to be reduced by hall.
After n half lives:
The amount remaing is:
[tex]A(n) = A(0)(0.5)^{n}[/tex]
In which A(0) is the initial amount and n is the number of half-lives.
Starting with 210 grams of a radioactive isotope, how much will be left after 6 half-lives?
This is A(6) when A(0) = 210. So
[tex]A(n) = A(0)(0.5)^{n}[/tex]
[tex]A(6) = 210(0.5)^{6} = 3.3[/tex]
Rounding to the nearest gram
3 grams will be left after 6 half-lives
Find a solution to the linear equation 9x+4y=−36 by filling in the boxes with a valid value of x and y.
Answer:
Please look at the picture below!
Step-by-step explanation:
Hope this helps!
If you have any question, please feel free to ask any time.
Please help with 4d.
Answer:
(Hemingway, The Old Man and the Sea)(Orwell, 1984)Step-by-step explanation:
A short web search will turn up the authors of the given titles:
The Old Man and the Sea - Hemingway
Huckleberry Finn - Twain
Moby D.ick - Melville
1984 - Orwell
Crime and Punishment - Dostoevsky
Multiply (2.0 ⋅ 10−4) ⋅ (3.1 ⋅ 10−20). Express the answer in scientific notation. 6.2 ⋅ 10−80 6.2 ⋅ 10−24 6.2 ⋅ 1024 6.2 ⋅ 1080
Answer:
6.2* 10 ^-24
Step-by-step explanation:
(2.0 ⋅ 10−4) ⋅ (3.1 ⋅ 10−20)
Multiply the numbers
2.0 * 3.1 =6.2
Add the exponents
10 ^-4 * 10 ^-20 = 10 ^( -4+-20) = 10 ^ -24
Put back together
6.2* 10 ^-24
The number is in scientific notation since there is one nonzero digit in front of the decimal
Answer:
B, now give the other person brainlist
A record club has found that the marginal profit, Upper P prime (x ), in cents, is given by Upper P prime (x )equals negative 0.0008 x cubed plus 0.35 x squared plus 45.5 x for x less than or equals 400, where x is the number of members currently enrolled in the club. Approximate the total profit when 240 members are enrolled by computing the sum
Answer:
The total profit when 240 members are enrolled is:
3,587,212.8 cents
Step-by-step explanation:
First of all, the total profit function is gotten by integrating the marginal profit function.
Integrate thus:
Total Profit = P(x) = [-0.0008x^4 ÷ 4] + [0.35x^3 ÷ 3] + [45.5x^2 ÷ 2]
P(x) = 0.0002x^4 + 0.1167x^3 + 22.75x^2
Next, substitute 240 for x, in the total profit function.
P(x) = 0.0002[240]^4 + 0.1167[240]^3 + 22.75[240]^2
P(x) = 663552 + 1613260.8 + 1310400
P(x) = 3,587,212.8 cents
Equivalent to $35,872.128
Which expression is equivalent to x^-5/3?
Answer:
[tex]\frac{1}{(\sqrt[3]{x} )^5}[/tex]
Step-by-step explanation:
[tex]x^{-\frac{5}{3} }[/tex] = [tex](\sqrt[3]{x} )^{-5}[/tex] = [tex]\frac{1}{(\sqrt[3]{x} )^5}[/tex]
Find the area of the shaded region if the dimensions of the unshaded region are 14ft x 18ft . Use 3.14 for π as necessary. Answer Asap Please! That would be greatly appreciated! PLEASE HELP ME ON THIS ASAP FIRST ANSWER GETS BRAINLIEST
Answer:
867.44 ft²
Step-by-step explanation:
The area of the shaded region is A = 196π + 252.
We have the dimensions of the unshaded region are 14ft x 18ft.
We have to find the area of shaded region.
What is the area of a Rectangle and a Circle?The area of a rectangle is -
A(R) = Length x Breadth = L x B
and the area of Circle is -
A(C) = [tex]\pi r^{2}[/tex]
According to the question -
Dimensions of the unshaded region -
L = 18ft
B = 14ft
Area of the shaded region (A) = Total Area - Area of Rectangle
Total Area = Area of 2 semicircles of radius (7 + 7) 14ft + Area of rectangle of length 18ft and breadth 28ft.
Total Area = ( [tex]2\times \frac{1}{2}\times \pi \times14 \times 14[/tex] ) + ( 18 x 28)
Total Area = 196π + 504
Area of the shaded region (A) = 196π + 504 - 252 = 196π + 252
Hence, the area of the shaded region is A = 196π + 252.
To solve more questions on Area of Figures, visit the link below -
https://brainly.com/question/9720037
#SPJ2
Which of the following points is a solution of the inequality y <-Ixl
You did not give any options but i will try to answer.
y < -lxl basically means that the value of y is less than the absolute value of x time - 1.
So if x = 2, then y is any number less than -2.
And if x is -3. then y is any number less than -3.
Happy to help!
ghvhhhhcycycugvuviggggggvvvggfgffgfggyg
Answer:
fortnite
Step-by-step explanation:
winwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwinwin
Answer:
Yes
Step-by-step explanation:
Yes
A pool that is 2.9 m tall cast a shadow that is 1.76 m long. At the Same time, a nearby building casts a shadow that is 38.25m long how tall is the building? round your answer to the nearest meter
Answer:
63m
Step-by-step explanation:
A pool that is 2.9m tall cast a shadow that is 1.76m
At the same time, a nearby building casts a shadow that is 38.25m long.
We are to find the height of the building.
If an object of length 2.9m cast an image of 1.76m ,
Then an image of 38.25m willl be cast by an object of what length?
Cross multiplying this gives:
[tex]\frac{38.25 * 2.9}{1.76}[/tex] = 63.02556818 = 63m (rounded up to nearest meter)
Graph the equation y = 1/2x + 5 on the coordinate plans provided below.
Answer:
Well, of course, i can not graph it in the coordinate plane in the image, but i can try to teach you how to do it.
We have the function y = (1/2)*x + 5
The first step is to generate pairs (x, y)
You can do it by evaluating the function in different values of x.
For example;
if x = 0
y = (1/2)*0 + 5 = 5
then we have the point (0,5)
if x = 2
y = (1/2)*2 + 5 = 6
Then we have the point (1, 6)
Now, with only two points you can graph the line.
First find the points in the coordinate plane, and mark them.
Now, with a ruler, draw a line that connects the two points.
That line is the graph of the function.
An example of the graph is:
The answer choices below represent different hypothesis tests. Which of the choices are one-tailed tests? Select all correct answers. Select all that apply: H0:p=0.46, Ha:p<0.46 H0:p=0.34, Ha:p≠0.34 H0:p=0.63, Ha:p≠0.63 H0:p=0.35, Ha:p≠0.35 H0:p=0.39, Ha:p<0.39
Answer:
H0:p=0.46, Ha:p<0.46
H0:p=0.39, Ha:p<0.39
Step-by-step explanation:
A one tailed test occurs in such a way that the value/results gotten is one sided and can either be lesser or greater than the particular given value but cannot be both.
Thus, in this case a one sided test includes
H0:p=0.46, Ha:p<0.46
H0:p=0.39, Ha:p<0.39
A rectangular vegetable garden will have a width that is 4 feet less than the length and an area of 140 square feet if x represents the length then the length can be found by solving the equation:
If x represents the length, then the length can be found by solving the equation: x(x-4)=140
Answer: x(x-4) = 140 This equation can be solve to find the length.
Step-by-step explanation:
If the width of the rectangle will be 4 less than the length and the length is represented by x then we could have the equation w=x - 4 for the width and the length is just x. And to find the area of a rectangle we multiply the length by the with so multiply x-4 by x for it to equal 140 because 140 is the area.
x(x-4) = 140 solve for x
[tex]x^{2}[/tex] - 4x = 140 subtract 140 from both sides
x^2 - 4x - 140 = 0 find a number that the product is 140 and the sum is -4
-14 and 10 works out.
[tex]x^{2}[/tex] - 14x + 10x - 140 = 0 factor by grouping
x(x -14) 10(x-14) = 0 factor out x-14
(x-14) ( x+10) = 0 set them both equal zero.
x-14= 0 or x+10 = 0
x = 14 or x= -10
Since -10 can't represent a distance, the answer is 14.
Check.
In this case the length is 14 and if the width is 4 less than the length then we will subtract 4 from 14.
14- 4 = 10 so the width is 10.
14 * 10 = 140
The perimeter of an equilateral triangle with a side length of 12 units is:
Answer:
Hey there!
An equilateral triangle has all sides equal to each other, so the perimeter would be 3x, where x is the length of one side.
Thus, the perimeter for this equilateral triangle would be 3(12)=36
Hope this helps :)
Answer:
[tex]\boxed{Perimeter = 36 \ units}[/tex]
Step-by-step explanation:
Perimeter = sum of all sides
Perimeter = 12 +12 + 12
Perimeter = 36 units
Suppose the results indicate that the null hypothesis should not be rejected; thus, it is possible that a type II error has been committed. Given the type of error made in this situation, what could researchers do to reduce the risk of this error? Choose a 0.01 significance level, instead of a 0.05 significance level. Increase the sample size.
Answer:
Increase the sample size.
Step-by-step explanation:
Increasing the sample size is the best way to reduce the likelihood of a type II error.
The type II error occurs when a hypothesis test accepts a false null hypothesis. That is, it fails to reject the null hypothesis that is false.
In such a situation, to increase the power of the test, you have to increase the sample size used in the test. The sampling size has the ability to detect the differences in a hypothesis test.
We have a bigger chance of capturing the difference if the sample size is larger, and it also increases the power of the test.
Find the equation of the line: parallel to 3x−y=11 through (−2, 0).
Answer:
[tex]\huge\boxed{y=3x+6\to 3x-y=-6}[/tex]
Step-by-step explanation:
[tex]\text{Let}\ k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ ||\ l\iff m_1=m_2\\k\ \perp\ l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\===========================[/tex]
[tex]\text{We have the equation of a line in the standard form:}\ 3x-y=11.\\\text{Convert it to the slope-intercept form:}\ y=mx+b\\\\3x-y=11\qquad\text{subtract}\ 3x\ \text{from both sides}\\-y=-3x+11\qquad\text{change the signs}\\y=3x-11\to \boxed{m_1=3}\\\\\text{Let}\ k:y=3x-11\ \text{and}\ l:y=mx+b.\\\\l\ ||\ k\Rightarrow l:y=3x+b\\\\\text{Substitute the coordinates of the point}\ (-2;\ 0)\ \text{to the equation of}\ l:\\\\0=3(-2)+b\\0=-6+b\qquad\text{add 6 to both sides}\\6=b\to\boxed{b=6}[/tex]
[tex]\text{Finally}\ l:y=3x+6\\\\\text{Convert to the standard form:}\\\\y=3x+6\qquad\text{subtract}\ 3x\ \text{from both sides}\\-3x+y=6\qquad\text{change the signs}\\l:\ 3x-y=-6[/tex]
Please answer this correctly without making mistakes
Answer:
Step-by-step explanation:
2.8 kilometers farther. Subtract 12.1km for Winchester and 9.3 for Stamford to get 2.8 kilometers.
mark wants to invest $10,000 for his daughter’s wedding. Some will go into a short term CD that pays 12% and the rest in a money market savings account that pays 5% interest. How much should he invest at Each rate if he wants to earn $1095.00 in interest in one year.
Consider three boxes containing a brand of light bulbs. Box I contains 6 bulbs
of which 2 are defective, Box 2 has 1 defective and 3 functional bulbs and Box 3
contains 3 defective and 4 functional bulbs. A box is selected at random and a bulb
drawn from it at random is found to be defective. Find the probability that the box
selected was Box 2.
Answer:
1/6
Step-by-step explanation:
As we already know that selected bulb is defective the required probability doesn't depend on functional bulbs at all.
The probability, that selected defective bulb is from Box2 is number of defective bulbs in Box 2 divided by total number of defective bulbs.
P(defective in box 2)= N(defective in box 2)/N(defective total)
As we know there is only 1 defective lamp in box 2.
So N(defective in box 2)=1
Total number of defective bulbs is Box1- 2 defective bulbs, box2- 1 defective bulbs, box3 - 3 defective bulbs. Total are 6 defective bulbs.
So N(defective total)=6
So P(defective in box 2)=1/6
Factor.
-7x2 + 49x
-7(x - 7)
01 - 7x2 + 49x)
-x(x – 7)
-7x(x – 7)
Answer:
Option 1: -7x (x - 7)
Step-by-step explanation:
Factor -7x out of -7x^2.
= -7x (x) + 49x
Factor -7x out of 49x.
= -7x (x) -7x (-7)
Factor -7x out of -7x (x) - 7x (-7).
= -7x (x - 7)
(Also I took the test and got this answer right)
Datguy323 is going to complain again. What's the variables for: [tex]x^2+y^2=29\\x+y=7[/tex]
y<4
Answer: :o I FINALLY MADE IT
(5, 2)
x = 5
y = 2
Step-by-step explanation:
First, I graphed both equations. They meet at the points (5,2) and (2,5). Because y < 5, the solution is (5, 2)
Hope it helps <3
Answer:
[tex]x=5\\y=2[/tex]
Step-by-step explanation:
[tex]x^2 +y^2 =29[/tex]
[tex]x+y=7[/tex]
Solve for x in the second equation.
[tex]x+y=7[/tex]
[tex]x+y-y=7-y[/tex]
[tex]x=7-y[/tex]
Plug in the value for x in the first equation and solve for y.
[tex](7-y)^2 +y^2 =29[/tex]
[tex]y^2-14y+49+y^2 =29[/tex]
[tex]2y^2-14y+20=0[/tex]
[tex]2(y-2)(y-5)=0[/tex]
[tex]2(y-2)=0\\y-2=0\\y=2[/tex]
[tex]y-5=0\\y=5[/tex]
[tex]y<4[/tex]
[tex]y=2[/tex]
[tex]y\neq 5[/tex]
Plug y as 2 in the second equation and solve for x.
[tex]x+y=7[/tex]
[tex]x=7-y[/tex]
[tex]x=7-2[/tex]
[tex]x=5[/tex]
A cone has a diameter of 8 units and a height of 8 units. Its radius is 4 units, and its volume is cubic units. A cylinder with the same height and radius as the cone will have a volume of cubic units. If a sphere has the same radius as the cylinder, its volume is the volume of the cylinder.
The above question is not complete because it was not written and arranged properly
Complete Question
1) A cone has a diameter of 8 units and a height of 8 units. Its radius is 4 units, and its volume is ______ cubic units.
2) A cylinder with the same height and radius as the cone will have a volume ______ of cubic units.
3) If a sphere has the same radius as the cylinder, its volume is ______the volume of the cylinder.
Answer:
1) Volume of the cone = 134.04cubic units
2)Volume of the cylinder = 402.12cubic units
3) Volume of the sphere= 268.08 cubic units. Hence, if a sphere has the same radius as the cylinder, its volume is 2/3 times the volume of the cylinder.
Step-by-step explanation:
1) A cone has a diameter of 8 units and a height of 8 units. Its radius is 4 units, and its volume is ______ cubic units.
Volume of a cone = 1/3πr²h
h = 8 units
r = 4 units
Volume = 1/3 × π × 4² × 8
134.04cubic units
2) A cylinder with the same height and radius as the cone will have a volume ______ of cubic units.
Volume of a cylinder = πr²h
Height and radius is the same as that of the cones hence,
h = 8 units
r = 4 units
= π × 4² × 8
= 402.12cubic units.
3) If a sphere has the same radius as the cylinder, its volume is ______the volume of the cylinder.
Volume of a Sphere = 4/3πr³
r = radius of the cylinder = 4 units
Volume of a Sphere = 4/3 × π × 4³
= 268.08 cubic units.
From the above question, we are asked to compare the volume of the sphere with the volume of the cylinder
Volume of the sphere : Volume of the cylinder
268.08 cubic units : 402.12 cubic units
268.08/402.12 = 2/3
Therefore, the volume of the sphere is 2/3 times the volume of the cylinder
In the equation, the value of a is:
Answer:
Please check if the answer is a = 4 or not
The half-life of a radioactive isotope is the time it takes for a quantity of the Isotope to be reduced to half its initial mass. Starting with 210 grams of a
radioactive isotope, how much will be left after 6 half-lives?
Round your answer to the nearest gram
Answer:
after 6 half lives: 210(1/2)^6= 3.28125
Step-by-step explanation:
isotope to be reduced to half its initial mass at first:
210(1/2)=105 half it is original weight
after second life: 210(1/2)^2=105(1/2)=52.5
after third : 210(1/2)^3=52.5/2=26.25
after fourth : 26.25/2=12.125
after fifth : 13.125/2
after 6 half lives: 210(1/2)^6= 3.28125
The smaller of two numbers is one-half the larger, and their sum is 27. Find the numbers. Answer: The numbers are ___ ___ ___
Answer:
the smaller is 9 while the digger is 18
The graph of an absolute value function has a vertex of (2,3) and crosses the x-axis at (−1,0) and (5,0). What is the equation for this absolute value function when y=0? A 0=|x+2|+3 B 0=|x−2|+3 C 0=−|x+2|+3 D 0=−|x−2|+3
Answer:
Option D.
Step-by-step explanation:
The vertex form of an absolute function is
[tex]y=a|x-h|+k[/tex]
where, a is a constant, (h,k) is vertex.
It is given that, vertex of an absolute function is (2,3). So, h=2 and k=3.
[tex]y=a|x-2|+3[/tex] ...(1)
It crosses the x-axis at (5,0). So put x=5 and y=0 to find the value of a.
[tex]0=a|5-2|+3[/tex]
[tex]-3=3a[/tex]
[tex]-1=a[/tex]
Put a=-1 in (1).
[tex]y=(-1)|x-2|+3[/tex]
[tex]y=-|x-2|+3[/tex]
Now, put y=0, to find the equation for this absolute value function when y=0.
[tex]0=-|x-2|+3[/tex]
Therefore, the correct option is D.
Answer:
I got this question on my test and I answered D cause if you look up the graph it matches the question
Step-by-step explanation:
D 0=−|x−2|+3
What is the value of the rate of change when we put a glass of water at room temperature in the freezer for 15 minutes, what is its temperature at 5 minutes and then at 10 minutes
Answer:
Step-by-step explanation:
I do not understand your question very well but I think that the state of the glass of the water would be liquid and solid at the same time since the water would not freeze 100% and would be with small pieces of solid and some parts liquid, in terms of temperature, I think that at 5 minutes it would be 5 ° C and at 10 it would be 3-4 ° C the longer it is in the freezer the less its temperature will be
(All of that is my point of view)
Average temperature 30 °
So :
30 ÷ 15 = 2 °
30 ÷ 10 = 3 °
30 ÷ 5 = 6 °