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2 years ago

FIND AC! PLS HELP!!

Mathematics

1 answer:

Anna007 [38]2 years ago

5 0

Answer:

AC = 7.3 in

Step-by-step explanation:

tan(51°) = 9/AC

1.24 = 9/AC

AC = 9/1.24

AC = 7.26

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Solve the equation for x. If necessary round to the nearest tenth.<br> x² = 17<br> X=<br> +

Ilya [14]

Answer:

x = 4.1 & x = −4.1

Step-by-step explanation:

$x^2 = 17\\$

x = ±√17

x = √17, −√17

x = 4.1 & x = −4.1

7 0

2 years ago

A top swimmer would be able to swim across a 25-meter pool in about 1/5

Anni [7]

Answer:

125 m/min

Step-by-step explanation:

Find the rate (speed) as follows:

25 meters

--------------------- = (125/1) meters/min = 125 m/min

(1/5) minute

This is equivalent to about 380 ft/min

7 0

2 years ago

A company that makes hair-care products had 8 comma 000 people try a new shampoo. Of the 8 comma 000 people, 16 had a mild all

Sedaia [141]

Answer:

around 0.2%

Step-by-step explanation:

the ratio is 16/8000, which divides out to 0.002, which as a percentage, is 2%

7 0

3 years ago

F(x)=x^2+16<br><img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E%7B2%7D%20%20%2B%2016" id="TexFormula1" title="f(x) = x^{2

deff fn [24]

$f(x) = {x}^{2} + 16 \\ \: \: \: \: \: \: \: \: \: \: \: (x - 4)(x + 4) \\ \: \: \: \: \: \: \: \: \: \: \: x = 4 \: \: \: \: \: x = - 4$

7 0

3 years ago

Find the minimum and maximum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f

nata0808 [166]

Via Lagrange multipliers:

$L(x,y,\lambd)=x^2y+x+y+\lambda(xy-5)$
$L_x=2xy+1+\lambda y=0$
$L_y=x^2+1+\lambda x=0$
$L_\lambda=xy-5=0$

$\underbrace{10}_{2xy}+1+\lambda y=0\implies \lambda=-\dfrac{11}y$
$xy=5\implies y=\dfrac5x\implies\lambda=-\dfrac{11}5x$

$\impliesx^2+1+\left(-\dfrac{11}5x\right)x=0\implies x^2=\dfrac56\implies x=\pm\sqrt{\dfrac56}$
$xy=5\implies y=\pm\sqrt{30}$

At these points, we get local minima of $f\left(\pm\sqrt{\dfrac56},\pm\sqrt{30}\right)=\pm2\sqrt{30}$ .

- - -

Another way to do this is to make $f(x,y)$ a function independent of $y$ , which is made possible by the constraint.

$xy=5\implies y=\dfrac5x$
$\implies f(x,y)=f\left(x,\dfrac5x\right)=F(x)=6x+\dfrac5x$
$\implies F'(x)=6-\dfrac5{x^2}=0\implies x=\pm\sqrt{\dfrac56}$

and so on.

8 0

3 years ago