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4 years ago

MARKING BRAINIEST HURRY UP****

Mathematics

1 answer:

ivann1987 [24]4 years ago

8 0

Answer:

Neither the scene nor the positions of the cameras are shown....

Without more detail of the problem, it would seem three different cameras would show the same scene only if they were all in exactly the same place, which is not possible.

Step-by-step explanation:

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7th grade math plss helpp

ddd [48]

Answer:

2x-2=-4

5/8x3/5= 15/40 then u simplify and get -4 3/8

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3 years ago

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PLZ answer this question y-5 ≥ -15

aalyn [17]

Answer:

y (is less than or equal to) 3

Step-by-step explanation:

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3 years ago

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-2(-2x + 3)-x -1 = -28

vekshin1

Answer:

x = -1

Step-by-step explanation:

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3 years ago

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X=1/16y^2 the directrix of the parabola is

yan [13]

To solve this problem you must apply the proccedure shown below:

1. You have the following equation of a parabola, given in the problem above:

x<span>=1/16y^2

2. Then, based on the graph attached, you have:

p=y^2/4x
p=8^2/(4)(4)
p=64/16
p=4

3. The directrix is:

directrix=h-p
directrix=0-4
directrix=-4

The answer is:-4</span>

3 0

3 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

<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

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

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

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

$x=0$ ----> 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

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

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