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9966 [12]
2 years ago
7

If f(x) = 4x²-2 then f(-2) is​

Mathematics
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} }

6 0
3 years ago
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Which one is it? this is so hard
Stels [109]

Answer:

attached

Step-by-step explanation:

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5 0
3 years ago
Please help me! 6y - 8x = 54
3241004551 [841]

Answer:

y = 9 + 4x/3

x = -27/4 + 3y/4

*everything that looks like x/y is a fraction, -27/4 is entirely negative

6 0
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

Step-by-step explanation:

15*5=75

75*5=375

375+ 25*10=625

792-625=167

8 0
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
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