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3 years ago

When the distributive property is used to solve the equation 10(x+8) =3x-3 what is the next step?

Mathematics

2 answers:

Alex73 [517]3 years ago

8 0

Answer:

multiply 10 and x+8

Step-by-step explanation:

gayaneshka [121]3 years ago

3 0

Answer:

$x=-\frac{83}{7}$

Step-by-step explanation:

use distributive property:

$10x+80=3x-3$

you have to move all the terms with $x$ to the left side of the equation:

subtract 3x from both sides

$10x+80-3x=-3$

combine like terms

$7x+80=-3$

subtract 80 from both sides

$7x=-3-80$

$7x=-83$

divide each term by 7

$\frac{7x}{7} =\frac{-83}{7}$

$x=-\frac{83}{7}$

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SOLVE THE QUESTION BELOW ASAP

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Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

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h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

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$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

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$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

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$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

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The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

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For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

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$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

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$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

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$82\ ft-34\ ft=48\ ft$

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

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The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

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