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3 years ago

I don't like math it makes me sad

Mathematics

2 answers:

PSYCHO15rus [73]3 years ago

8 0

It would be 250 in because you see 10x20 is 200 and and 10x10 is 100. But since part of the figure is a triangle you have to dive it by 2 and 100/2=50. 200+50=250
Hope that helps!

Irina18 [472]3 years ago

7 0

Answer:

300 $in^{2}$

Step-by-step explanation:

First, calculate the area of the rectangle. The height and width are 10 and 20, so the area is 200 $in^{2}$ .

For the triangle, we know the base length is 10 inches. But for the height, you add the 10 inches that happens to be part of the rectangle and another 10 inches, which is shown to the left of the top of the triangle, making the height 20 inches. Following the formula for the area of a triangle, $\frac{bh}{2}$ , you would get 100 $in^{2}$ .

To get the total area of the whole figure, add 200 and 100, which will give you 300 $in^{2}$

also math makes me sad sometimes too so that's very relatable

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The diagonal of a rectangle is 25 in. The width is 15 in. What is the area of the rectangle?

Nana76 [90]

Answer:

375 in² is the answer.

Step-by-step explanation:

Are of rectangle= area of 2 triangle

= 1/2*25*15+1/2*25*15

=187.5+187.5

=375 in²

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2 years ago

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slava [35]

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If the radius of a circle with a area of 5.3 mm squared is multiplied by 5,

chubhunter [2.5K]

I'm guessing the question is meant to be "If the radius of a circle with a area of 5.3 mm squared is multiplied by 5, what will the new area be?", so i'll answer that.

Area of a circle =πr^2

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3 years ago

Identify each function that has a remainder of -3 when divided x+6

Sergeu [11.5K]

Answer:

Step-by-step explanation:

According to remainder theorem, you can know the remainder of these polynomials if you plug in x = -6 into them.

So we will plug in -6 into x of all the polynomials ( A through D) and see which one equals -3.

For A:

$x^5 + 2x^2 - 30x + 30\\=(-6)^5 + 2(-6)^2 - 30(-6) + 30\\=-7494$

For B:

$x^4 + 4x^3 - 21x^2 - 53x + 12\\=(-6)^4 + 4(-6)^3 - 21(-6)^2 - 53(-6) + 12\\=6$

For C:

$x^3 - 10x^2 - 7\\=(-6)^3 - 10(-6)^2 - 7\\=-583$

For D:

$x^4 + 6x^3 - 10x - 63\\=(-6)^4 + 6(-6)^3 - 10(-6) - 63\\=-3$

The only function that has a remainder of -3 when divided by x + 6 is the fourth one, answer choice D.

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3 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

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3 years ago