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3 years ago

Find the value of x

Mathematics

1 answer:

ElenaW [278]3 years ago

6 0

Answer:

$\large\boxed{x\approx12.94}$

Step-by-step explanation:

Use cosine:

$\cosine=\dfrac{adjacent}{hypotenuse}$

We have:

$adjacent=4\\hypotenusex\\\alpha=72^o\\\\\cos72^o\approx0.309$

Substitute:

$\dfrac{4}{x}=0.309$

$\dfrac{4}{x}=\dfrac{309}{1000}$ <em>cross multiply</em>

$309x=4000$ <em>divide both sides by 309</em>

$x\approx12.94$

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What single decimal multiplier would you use to increase by 4% followed by a 3% increase

Mandarinka [93]

Answer:

1.0712

Step-by-step explanation:

if you increase by 4%, it means you will have a new total of 104%. In decimals it is 1.04 (divide your percentage by 100) after that you do the same with the 3% increase. you'll have a new total of 103%. In decimals 1.03. If you want to know what single multiplier you can use, you need to multiply the both decimals. So you get 1.04 x 1.03 = 1.0712

hope this will help you, have a nice day!

7 0

4 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B%5Cfrac%7B5%7D%7B6%7D%20%7D%20%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%20%7D"

Kryger [21]

Answer:

x^2/3

Step-by-step explanation:

(x^5/6)/(x^1/6)=x^(5/6-1/6)=x^4/6

simplify 4/6, you get 2/3.

4 0

3 years ago

It costs a clothing company $463.45 to make 169 T-shirts.

Artemon [7]

Answer:

The unit price for 1 T-shirt is about $2.74.

Step-by-step explanation:

Set up an equation:

Variable x = unit price for 1 T-shirt

169x = 463.45

Divide both sides by 169:

x = 2.742307...

Round to the nearest cent:

x = 2.74

3 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is