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2 years ago

7

If f(x) = 4x²-2 then f(-2) is

1 answer:

klemol [59]2 years ago

3 0

To find f(-2), substitute (-2) into x of the function f(x) = 4x^2-2
We get, f(-2) = 4(-2)^2 - 2
= (4*4)-2 = 16-2 = 14
Therefore, f(-2)=14.

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Rewrite the expression with rational exponents as a radical expression. 7 times x to the two thirds power

Daniel [21]

Answer:

$(7x)^{\frac{2}{3}}$ = $\sqrt[3]{49x^{2} }$

Step-by-step explanation:

We have been given the expression;

7 times x to the two thirds power which can be written mathematically as;

$(7x)^{\frac{2}{3}}$

To express the above expression as a radical, we need to recall that;

$a^{\frac{b}{n}}=\sqrt[n]{a^{b}}$

Therefore;

$(7x)^{\frac{2}{3}}=\sqrt[3]{(7x)^{2} }\\\\=\sqrt[3]{49x^{2} }$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Please help me! 6y - 8x = 54

3241004551 [841]

Answer:

y = 9 + 4x/3

x = -27/4 + 3y/4

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3 years ago

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Walker is reading a book that is 792 pages. He reads 15 pages a day during the week, and 25 pages a day during the weekend. Afte

liubo4ka [24]

Answer:

167 Pages Left To Read

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3 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago