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3 years ago

A quadratic equation is shown below:

Mathematics

1 answer:

AfilCa [17]3 years ago

5 0

the solutions to a general quadratic equation is

X=-b±√b²-4ac/2a ,, When ax²+bx+c=0

the discriminant is the expression under the radical b²−4ac

Part A:

discriminant is (-16)² - 4(9)(60) = -1904

there are two complex solutions

Part B:

4x² + 8x − 5 = 0

4x² + 10x -2x - 5=0

2x ( 2x + 5) - 1(2 x + 5) =0

(2x + 5)(2x-1) = 0

/ \

2x+5 = 0 2x-1 =0

x = -5/2 x = 1/2

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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

2 years ago

Giving brainliest and alot of points! A projectile is launched at an angle (θ) of 36° and travels a distance of 375 feet; gravit

abruzzese [7]

Answer:

Step-by-step explanation:

4 = 112.33 ft/sec = 112 ft/sec

6 0

1 year ago

Find the product -7(-a2)(-b3)

vlada-n [284]

$-7(-a^2)(-b^3)=\boxed{-7a^2b^3}$

7 0

3 years ago

If the system given has parallel lines, what is the answer?

uysha [10]

We need to see the "given system".

7 0

2 years ago

Read 2 more answers

4y+12=0<br> How do I get the two points to be graphed

coldgirl [10]

Did somebody say you're supposed to draw the graph of the equation ?
Is that the assignment ?

OK. Just like every other equation you need to graph, get it in the
standard form, where 'y' is all alone on one side, and everything else
is on the other side. When you do that, you'll be able to spot the slope
and y-intercept of the line, or get some points, or whatever you want.

4y + 12 = 0

Subtract 12 from each side: 4y = -12

Divide each side by 4: y = -3

There's the equation you can handle.
The y-intercept is -3, and the slope is zero.

Would you like some points ? OK. Pick a couple of values for 'x',
and calculate the value of 'y' for each one:

The first value I picked for 'x': x = 72
The equation is y=-3, so when x=72, y=-3. The point is (72, -3)

The second value I picked for 'x' is: x = 1
The equation is y=-3, so when x=1, y=-3. The second point is (1, -3).

The third value I picked for 'x' is 4 billion.
The equation is y=-3, so when x=4 billion, y=-3. The third point is (1, -3).

Do you see what's going on here ? Your original equation didn't even
have 'x' in it, so we could tell right away that when the graph is drawn,
the value of 'y' at every point can't depend on 'x'.

When we simplified the equation and got it in standard form, we found that
the slope of the graph is zero. That means the graph doesn't rise or fall ...
it's just a horizontal line. Sure enough, the height of points on the line
doesn't depend on 'x'. The value of 'y' at every point on the line is -3 .

8 0

3 years ago