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3 years ago

Find the approximate side length of a square game board with an area of 161 in?

1 answer:

4 0

Answer: 12.69 in

Explanation:
Square root of 161 is around 12.69
That means each side of the square is about 12.69 bc l x w = a
and square root of a = side length of square

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The value of the number 7 in 2, 710 is times the value of 7 in 1,070

Irina18 [472]

10 times. That's the answer :)

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4 years ago

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What percent of 60 is 12? Show your work. Is this wrong or right. 60/100 = 0.6*12 7.2% or do i removes the .2 off the percentage

Troyanec [42]

You don't just go removing numbers here and there without a good reason. In any case, your whole approach here is wrong. You're calculating 60 percent of 12. But the question says to take 12 out of a group of 60 and tell what percent of the group you have. That's (12/60) x 100.

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4 years ago

In Orlando, Florida, the high temperature was 70°F on Saturday and 98°F on Sunday. What is the percent of change in the high tem

dimaraw [331]

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3 years ago

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ABCD is a rectangle. The measure of angle BCE is 4x-23 and the angle measure of angle DCE is 5+5x. Find the measure of angle BEC

Temka [501]

<h2>Explanation:</h2><h2></h2>

The diagram for this problem is shown below. Point E is the intersection of the two diagonals and we know that:

$\angle BCE=\beta \\ \\ \angle DCE=\theta$

Every internal angle of any rectangle measures 90 degrees, so:

$\beta +\alpha=90 \\ \\ (4x-23)+(5+5x)=90 \\ \\ \\ Isolating \ x: \\ \\ (4x+5x)+(5-23)=90 \\ \\ 9x-18=90 \\ \\ 9x=108 \\ \\ x=\frac{108}{9}=12$

So:

$\beta=4x-23=4(12)-23=25^{\circ}$

So the measure of angle BEC can be found as follows:

We know that triangle ΔCEB is an isosceles triangle because the diagonals of any rectangle measure the same. So

$\angle EBC=\beta$

The sum of internal angles of any triangle add up to 180 degrees, so:

$\beta + \beta+\angle BEC=180^{\circ} \\ \\ 2\beta+\angle BEC=180^{\circ} \\ \\ \angle BEC=180^{\circ}-2\beta \\ \\ \angle BEC=180^{\circ}-2(25^{\circ}) \\ \\ \angle BEC=180^{\circ}-50^{\circ} \\ \\ \boxed{\angle BEC=130^{\circ}}}$

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4 years ago

Why can you NOT combine 3x and 5x^2 ?

diamong [38]

Although they are the same variable, you cannot add two variables raised to different powers. 3x^1 + 5x^2 cannot work, but 3x^2 + 5x^2 can. Also, if you are multiplying them, they would combine to be 15x^3, as 3 and 5 (The coefficients) multiply together and x^2 times x^1 = x^3.

I hope that helps.

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4 years ago