All categories

3 years ago

How can you find f(3) if f(x) = -2x2 – 4?

Mathematics

2 answers:

Julli [10]3 years ago

8 0

Answer: f(3)=-20

Step-by-step explanation:

If you look at f(x) and f(3), you can see that 3 is in place of x. That means when x=3, what is f(x). You plug in x=3 into f(x) and solve.

f(3)=-2(3)²-2

f(3)=-2(9)-2

f(3)=-18-2

f(3)=-20

Citrus2011 [14]3 years ago

8 0

Answer:

-20

Step-by-step explanation:

You might be interested in

What is the average of 4741, 4261, 4628?

lara [203]

Answer:

4628

I hope it will help you all

8 0

3 years ago

Is 3/16 equivalent to 3/8? How do you know?

Ymorist [56]

Answer: 3/16 is not equivalent to 3/8

Step-by-step explanation: 3/16 would be like eating 3 pieces of a pie that has 16 pieces whilst 3/8 would be like eating 3 pieces of a pie that has 8 pieces. They are mildly similar but 3/16 is twice as small as 3/8 and henceforth not equivalent.

7 0

3 years ago

Read 2 more answers

You have a 36 foot fence that have a 4 equal side you painted 1/4 what fraction have you completed

k0ka [10]

we have completed 9/36=1/4 of the fence

5 0

3 years ago

Can anyone help with math models?? Please…?

Solnce55 [7]

Answer:

12*1/4= 3

30 is C

16*3/4 = 12

31 is B

8 0

2 years ago

Read 2 more answers

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago