Equivalent Fractions

Write the expression in complete factored form. 3b(p + 2) - 7(p + 2) =

pshichka [43]

If you factor (p+2) out of the equation, this leaves (3b-7), so the final answer is (p+2)(3b-7)

5 0

3 years ago

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Find the equation of the tangent to the circle x2 + y2 = 109 at point (-10,3)

konstantin123 [22]

Answer:

10x - 3y = -109 .

Step-by-step explanation:

x^2 + y^2 = 109

Implicit Differentiation:

2x + 2y.y' = 0

y' = -2x/2y = -x/y

So at the point (-10,3) the gradient of the tangent is -(-10)/3 = 10/3.

Equation of the tangent:

y - y1 = m(x - x1)

y - 3 = 10/3(x + 10)

y - 3 = 10/3 x + 100/3

3y - 9 = 10x + 100

-109 + 3y = 10x

10x - 3y = -109

4 0

3 years ago

For school spirit day, 11.875% of your class wears orange shirts, 5/8 of your class wears blue shirts, 0.15625 of your class wea

Rzqust [24]

Answer:

Gold, Orange, White, Blue

Step-by-step explanation:

8 0

3 years ago

For a fundraiser, Jose sold a total of 111 Snickers and Twix bars.

elena55 [62]

Answer:

hello

Step-by-step explanation:

hello

3 0

4 years ago

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....

$0=-16t^2+500t+20\\ \\ -4(4t^2-125t-5)=0\\ \\ 4t^2-125t-5=0$

Using quadratic formula, we will get.....

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(-5)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{15705}}{8}\\ \\ t=31.2899...\\ and\\ t= -0.0399...$

(Negative value is ignored as time can't be in negative)

So, the rocket will remain in the air for 31.2899... seconds.

5 0

4 years ago