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2 years ago

9

Please help me with this math problem (image below)

1 answer:

VashaNatasha [74]2 years ago

6 0

Answer:

Option B.

Step-by-step explanation:

Obvious. <u>Lets explain:</u>

<em>Option A: It's pretty ease to do. You just need 2 triangles to make a parallelogram, and that's that.</em>

<em>Option B: How can you make a rectangle from triangles-???</em>

<em>Option C: You might think I am wrong, but I am right. Get 6 triangles, then plato, you have a whole hexagon!</em>

<em>Option D: You probably already know this...but you just need three triangles.</em>

It's correct on every school thing.

<h2>Option B is your answer!</h2>

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Janice and her dad are building a treehouse. Her dad bought a box of 155 nails for $3.10 and a box of 85 screws for $7.65. How m

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2 years ago

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The coordinates of the vertices of a rectangle are (−3, 4) , (7, 2) , (6, −3) , and (−4, −1) . What is the perimeter of the rect

ki77a [65]

Answrer

Find out the what is the perimeter of the rectangle .

To prove

Now as shown in the figure.

Name the coordinates as.

A(−3, 4) ,B (7, 2) , C(6, −3) , and D(−4, −1) .

In rectangle opposite sides are equal.

Thus

AB = DC

AD = BC

Formula

$Disatnce\ formula = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2}}$

Now the points A(−3, 4) and B(7, 2)

$AB = \sqrt{(7- (-3))^{2} +(2- 4)^{2}}$

$AB = \sqrt{(10)^{2} +(-2)^{2}}$

$AB = \sqrt{100+4}$

$AB = \sqrt{104}$

$AB = 2\sqrt{26}\units$

Thus

$CD= 2\sqrt{26}\units$

Now the points

A (−3, 4) , D (−4, −1)

$AD = \sqrt{(-4 - (-3))^{2} +(-1- 4)^{2}}$

$AD = \sqrt{(-1)^{2} +(-5)^{2}}$

$AD = \sqrt{1 + 25}$

$AD = \sqrt{26}\units$

Thus

$BC = \sqrt{26}\units$

Formula

Perimeter of rectangle = 2 (Length + Breadth)

Here

$Length = 2\sqrt{26}\ units$

$Breadth = \sqrt{26}\ units$

$Perimeter\ of\ rectangle = 2(2\sqrt{26} +\sqrt{26})$

$Perimeter\ of\ rectangle = 2(3\sqrt{26})$

$Perimeter\ of\ rectangle = 6\sqrt{26}$

$\sqrt{26} = 5.1 (Approx)$

$Perimeter\ of\ rectangle = 6\times 5.1$

Perimeter of a rectangle = 30.6 units.

Therefore the perimeter of a rectangle is 30.6 units.

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Step-by-step explanation:

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Veseljchak [2.6K]

For $x$ between $-\pi$ and $\pi$ , we have

$\cos x=\frac{\sqrt2}2=\frac1{\sqrt2}$ when $x=\pm\frac\pi4$ ;
$\cos x=1$ for $x=0$ ; and
$\cos x=0$ for $x=\pm\frac\pi2$

$\cos x$ is continuous over its domain, so the intermediate value theorem tells us that

$\cos x\ge\frac{\sqrt2}2$

is true for $-\frac\pi4\le x\le\frac\pi4$ .

For all $x$ , we take into account that $\cos x$ is $2\pi$ -periodic, so the above inequality can be expanded to

$-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4$

where $n$ is any integer. Equivalently,

$-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi$

To get the corresponding solution set for

$\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2$

simply replace $x$ with $x+\frac\pi3$ :

$-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi$

$\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}$

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2 years ago