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3 years ago

Solve the equation: -17-2x=6-x Solve for x.

Mathematics

2 answers:

lions [1.4K]3 years ago

8 0

Answer: x = -23

Step-by-step explanation:

-17-2x=6-x

+x +x

-17-x=6

+17 +17

-x=23

/-x /-x

x=-23

irinina [24]3 years ago

7 0

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Vedmedyk [2.9K]

Answer:

angle 4

Step-by-step explanation:

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2 years ago

The equation h(t) = -16t² + 80t + 64 represented the height, in feet, of a potato t seconds after it has been launched.

Lyrx [107]

Answer:

Part A) The potato hit the ground at t=5.70 seconds (see the explanation)

Part B) The potato is 40 feet off the ground at the time t=5.28 seconds (see the explanation)

Step-by-step explanation:

we have

$h(t)=-16t^2+80t+64$

where

h(t) is the height of a potato in feet

t is the time in seconds

Part A) Write an equation that can be solved to find when the potato hits the ground. Then solve the equation

we know that

When the potato hit the ground, the value of h(t) must be equal to zero

For h(t)=0

$-16t^2+80t+64=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^2+80t+64=0$

$a=-16\\b=80\\c=64$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(64)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{10,496}} {-32}$

$t=\frac{-80+\sqrt{10,496}} {-32}=-0.70$

$t=\frac{-80-\sqrt{10,496}} {-32}=5.70$

therefore

The potato hit the ground at t=5.70 seconds

Part B) Write an equation that can be solved to find when the potato is 40 feet off the ground. Then solve the equation

For h(t)=40 ft

substitute in the quadratic equation

$-16t^2+80t+64=40$

$-16t^2+80t+24=0$

Solve the quadratic equation

we have

$a=-16\\b=80\\c=24$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(24)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{7,936}} {-32}$

$t=\frac{-80+\sqrt{7,936}} {-32}=-0.28$

$t=\frac{-80-\sqrt{7,936}} {-32}=5.28$

therefore

The potato is 40 feet off the ground at the time t=5.28 seconds

3 0

2 years ago

Find the volume of each figure. Round your answers to the nearest tenth, if necessary

olga nikolaevna [1]

Answer:

1.92 m³

hopefully this answer can help you to answer the next question.

5 0

2 years ago

7-18 use part 1 of the fundamental theorem of calculus to find the derivative of the function.

blondinia [14]

$\displaystyle h(x)=\int\limits_{1}^{\sqrt{x}}~\cfrac{z^2}{z^4+1}dz~\hspace{10em}\cfrac{dh}{du}\cdot \stackrel{chain~rule}{\cfrac{du}{dx}\implies \cfrac{dh}{dx}} \\\\[-0.35em] ~\dotfill\\\\ u=\sqrt{x}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}} \\\\[-0.35em] ~\dotfill$

$\cfrac{dh}{dx}\implies \displaystyle \cfrac{d}{du}\left[ \int\limits_{1}^{u}~\cfrac{z^2}{z^4+1}dz \right]\cdot \cfrac{1}{2\sqrt{x}}\implies \left[ \cfrac{u^2}{u^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \stackrel{substituting~back}{\left[ \cfrac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}}}\implies \cfrac{x}{x^2+1}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{\sqrt{x}}{2x^2+2}$

7 0

1 year ago

How do I get my answer

kupik [55]

Answer:

i think it is 5/8 6/12 3/4

Step-by-step explanation:

5 0

3 years ago