The sum of three integers is 92. The second number is three times the first number. The third number is ten less than

On a basketball court the free throw line is marked off geometrically this area of the court is called a key and is topped

Alenkinab [10]

Step-by-step explanation:

Arc length formula is

$s = rx$

where x is radinas.

A semi circle has a radian measure of

$\pi$

The radius is half of the diameter, 12 so the radius is

$12 \times \frac{1}{2} = 6$

So the arc length is

$6 \times \pi = 6\pi$

Area of semi circle is

$\frac{1}{2} \pi {r}^{2}$

where r is the radius.

$\frac{1}{2} \pi6 {}^{2}$

$\frac{1}{2} 36\pi$

$18\pi$

3 0

3 years ago

Change the following into the form f(x)=mx+b a. 8x-y=2 b.-3x+2y=9

kiruha [24]

Answer:

B

Step-by-step explanation:

I took the test

4 0

3 years ago

SOMEONE HELP ME IM FREAKING OUT I LITERALLY CANT WITH THIS QUESTION IM PRAYING PLEASE HELP ME IM SO SERIOUS IM GONNA END IT PLS

antiseptic1488 [7]

Answer:

$\sf -11+7\sqrt{2}$

Step-by-step explanation:

Given expression:

$\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }$

Rewrite 32 as 16 · 2:

$\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }$

Apply radical rule $\sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}$

$\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }$

As $\sf \sqrt{16}=4$ :

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }$

Multiply by the conjugate:

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }$

$\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}$

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}$

As $\sf \sqrt{2}\sqrt{2}=\sqrt{4}=2$ :

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}$

$\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}$

$\sf \implies \dfrac{11-7\sqrt{2}}{-1}$

$\sf \implies -11+7\sqrt{2}$

7 0

2 years ago

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0

3 years ago

$2 increased by 8.5%

Taya2010 [7]

You multiply 2 by 0.085 and you get the answer
$0.17
Then add $0.17 and $2 and you get $2.17

8 0

3 years ago