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2 years ago

Write an expression for the area of the shaded region in its simplest form. Show all of your steps.

1 answer:

6 0

The area of the shaded region is going to be the area of the rectangle minus the area of the square.

Area of a rectangle is L * W.
A = L * W
A = (x + 10)(2x + 5)
A = x(2x + 5) + 10(2x + 5)
A = 2x^2 + 5x + 20x + 50
A = 2x^2 + 25x + 50 .....this is the area of the rectangle

area of a square is : A = a^2...where a is one side
A = (x + 1)^2
A = (x + 1)(x + 1)
A = x(x + 1) + 1(x + 1)
A = x^2 + x + x + 1
A = x^2 + 2x + 1

now we subtract the area of the square from the area of the rectangle to get the area of the shaded region.

2x^2 + 25x + 50 - (x^2 + 2x + 1) =
2x^2 + 25x + 50 - x^2 - 2x - 1 =
x^2 + 23x + 49 <== the area of the shaded region

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What is the measure of ACE in the diagram below? A. 46 B. 104 C. 29 D. 58

beks73 [17]

Answer:

Step-by-step explanation:

The secant- secant angle ACE is half the difference of the measures of the intercepted arcs, that is

∠ ACE = $\frac{1}{2}$ (AE - BD ) = $\frac{1}{2}$ (104 - 46)° = $\frac{1}{2}$ × 58° = 29° → C

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3 years ago

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Betty had $70. She bought a necklace for $9 and a ring for $7. Which expression correctly shows the total money Betty has left?

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70 - 9 - 7 = x 70 being the original amount

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2 years ago

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Given the points (2, 2) and (-1, -7) find the slope.

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Slope is change in y over the change in x.

Slope = (-7 - 2) / (-1 -2)

Slope = -9/-3

Slope = 3

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2 years ago

The common point between lines y = 2x + 5 and y = ½ x + 6 is (3, 1/2). true or false

mixer [17]

The common point between y = 2x + 5 and y = (1/2)x + 6 will have the same values for x and y.

Therefore, set the two y expressions equal to obtain
2x + 5 = (1/2)x + 6

Subtract (1/2)x from each side.
(3/2)x + 5 = 6

Subtract 5 from each side.
(3/2)x = 1

Multiply each side by 2/3.
x = 2/3.

From the first equation, obtain
y = 2*(2/3) + 5 = 19/3.

The common point is (2/3, 19/3). It is not equal to (3, 1/2).

Answer: False

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2 years ago

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A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. The average nicotine content from a

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Answer:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

$df=n-1=15-1=14$

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

Step-by-step explanation:

Information provided

$\bar X=42.6$ represent the average nicotine content

$s=3.7$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =40$ represent the value to check

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:

Null hypothesis: $\mu \leq 40$

Alternative hypothesis: $\mu > 40$

The statistic for this case would be:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replcing we got:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

The degrees of freedom are:

$df=n-1=15-1=14$

The p value would be:

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

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3 years ago