All categories

2 years ago

What is f(-2) for f(x)=1/2x^2

Mathematics

1 answer:

GarryVolchara [31]2 years ago

7 0

Answer: 2

Steps:
f(-2) = 1/2x^2
f(-2) = 1/2(-2)^2
f(-2) = 1/2(4)
f(-2) = 2

You might be interested in

Graph the inequality.<br> -6 ≤y + 2x < 15

omeli [17]

Inequalities help us to compare two unequal expressions. The correct option is B.

<h3>What are inequalities?</h3>

Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed. It is mostly denoted by the symbol <, >, ≤, and ≥.

The given inequality -6 ≤y + 2x < 15 can be broken into two small inequality, as shown below.

-6 ≤y + 2x
y + 2x < 15

Now, if we plot the inequality as shown below, then the area in which both the shaded region overlap is the area of the this inequality.

Hence, the correct option is B.

Learn more about Inequality:

brainly.com/question/19491153

#SPJ1

6 0

2 years ago

How do I solve this math problem?

Nat2105 [25]

Answer: 4x² + 4x - 2 - $\frac{3}{2x+3}$

<u>Step-by-step explanation:</u>

You can use either synthetic or long division. I am using synthetic division:

2x + 3 = 0 ⇒ x = $-\frac{3}{2}$

$-\frac{3}{2}$ | 4 10 4 -6

4 4 -2 -3 ← <em>-3 is the remainder</em>

↓ ↓ ↓

4x² +4x -2 ← <em>factored polynomial</em>

3 0

3 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

What slope is perpendicular to x + 4y = 12

myrzilka [38]

Solve for the equation. The. Divide by 4 on both sides. And whatever you get for the slope. Write the opposite which will give you the perpendicular slope. For example: 1/4x is perpendicular to -4x

6 0

3 years ago

Explain how a square is a recttangle

maw [93]

Answer:

a square is a rectangle as it's a shape with 4 sides with 90degree angles, and pairs of sides are equal.

4 0

3 years ago

Read 2 more answers