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Tasya [4]
3 years ago
7

If m = 5 and n = -2, evaluate 3m - 4n and n2 - m and find the product of the two expressions.

Mathematics
1 answer:
ddd [48]3 years ago
8 0

Hence the correct answer is C) -23

Further explanation:

We have to put the values of m and n into both expression first separately and then multiply the answers.

Given expressions are:

Exp\ 1:3m-4n\\Exp\ 2: n^2-m

The values are:

m=5\\n=-2

Putting the values in expression 1:

3m-4n\\=3(5)-4(-2)\\=15+8\\=23

Putting the values of m and n in second expression

n^2-m\\=(-2)^2-5\\=4-5\\=-1

Multiplying both expressions:

(3m-4n)(n^2-m)=23* (-1)=-23

Hence the correct answer is C) -23

Keywords: Polynomials, Putting values in expressions

Learn more about polynomials at:

  • brainly.com/question/4142886
  • brainly.com/question/4934417

#LearnwithBrainly

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Find the surface area of the part of the circular paraboloid z=x^2+y^2 that lies inside the cylinder X^2+y^2=1
hichkok12 [17]

Answer:

\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}

Step-by-step explanation:

Given that:

The surface area (S.A) z = x^2 +y^2

Hence the S.A is of form z = f(x,y)

Then the S.A can be represented with the equation

A(S) = \iint _D \sqrt{1+ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2} \ dA

where :

D = cylinder x^2 +y^2 =1

In polar co-ordinates:

D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)

Similarly, \dfrac{\partial z}{\partial x} = 2x and \dfrac{\partial z}{\partial y} = 2y

Therefore;

S.A = \iint_D \sqrt{1+4x^2+4y^2} \ dA

= \iint_D \sqrt{1+4(x^2+y^2)} \ dA

= \int^{2 \pi}_{0} \int^{1}_{0}  \sqrt{1+4r^2} \ r \ dr \d \theta

= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0

= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]

\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}

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ANSWER

1.39

EXPLANATION

The given quadratic equation is

0 = 2 {x}^{2}  + 3x - 8

This is the same as,

2 {x}^{2}  + 3x - 8 = 0

Comparing to

a {x}^{2}  + bx  + c = 0

We have

a=2, b=3,c=-8

Using the quadratic formula, the solution is given by:

x =  \frac{ - b \pm \:  \sqrt{ {b}^{2}  - 4ac} }{2a}

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