if the sum of a number and nine is tripled the results is seven less than twice the number. find the number

Since It's parallel to y=5x-9 the slope of this new line is 5
so the equation looks like y=5x+b
to find be we just replace x by 2 and y by 3
3=10-b so b = -7
the equation is y=5x-7

7 0

4 years ago

SOLVE THE QUESTION BELOW ASAP

qwelly [4]

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

Let

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

t -----> the time in seconds

we have

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

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

Find the vertex

The function is written in vertex form

so

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

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

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

square root both sides

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

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

therefore

The t-intercepts are

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

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

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

so

$x=0$ ----> 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

we know that

For t=0

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

$h(0)=98\ ft$

For t=1

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

$h(1)=82\ ft$

Find the difference

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

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

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

$h(1)=82\ ft$

For t=2

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

$h(2)=34\ ft$

Find the difference

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

so

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

therefore

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

8 0

3 years ago

I WILL GIVE BRAINLIEST!

Nesterboy [21]

Answer:

$6

Step-by-step explanation:

So we get 5 hrs.

1____2____3____4____5

$2 + $1 + $1 + $1 + $1 = $6

4 0

3 years ago