All categories

3 years ago

Identify the y-intercept 1.)v^2 - 11v+30=0 2.) v^2 + 2v - 8=0

1 answer:

6 0

We know that y-intercepts are the values found when x=0;

1) v²-11v+30=y <-- replace 0 as y and substitute x=0
(0)²-11(0)+30=y
30=y

The y-intercept is 30 here.

2) v²+2v-8=y <-- repeat the above steps
(0)²+2(0)-8=y
-8=y

The y-intercept is -8 here

As you can notice, the c value in the standard equation (ax²+bx+c=y) is the y-intercept.

Hope I helped :)

You might be interested in

Cornyia swims a 1/2 mile every 1/4 hour how long will it take to swim 2/ 1/2 miles

Pavlova-9 [17]

Answer:

I'm unsure what the final fraction is but if you comment it I'll reply with the answer

Step-by-step explanation:

every hour she travels 2 miles, so in 30 minutes she will travel 1 mile, hopefully that can help

5 0

3 years ago

There are 12 counters in a bag.

astraxan [27]

Answer:

number of green counters

=12-3+1+2

number of green and yellow counters

=6+2

probability of getting green or yellow counter

=8/12

=2/3

Step-by-step explanation:

5 0

2 years ago

Help help help ASAP

STatiana [176]

Answer:

x=-8

Step-by-step explanation:

Subtract 2 from both sides

Simplify

Multiply all terms by the same value to eliminate fraction denominators

Cancel multiplied terms that are in the denominator

Subtract 2x from both sides

Simplify

8 0

2 years ago

Read 2 more answers

Here is the Triangle <br><br> What is the measure of angle A?

Luba_88 [7]

Answer:

45 degree angle

Step-by-step explanation:

Pleeaase correct me if im wrong

3 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

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

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago