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noname [10]
3 years ago
5

PLEASE help me ! 10 points + brainliest :))

Mathematics
1 answer:
Yanka [14]3 years ago
7 0

Answer:

200

Step-by-step explanation:

Mark me as brainliest

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X/-6= -20. What is x?
lesya692 [45]
The answer is x = 120
6 0
3 years ago
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The equation of line / is y=3x/a +5, where a is a positive constant. If the value of a in this equation is doubled, then the res
Anettt [7]

The resulting equation will represent a line whose slope is 1/2 times the slope of the line

<h3>How to determine the slope of the new line?</h3>

The equation of the line is given as:

y = 3x/a + 5

The constant a is a positive constant.

So, when the value of a in the equation is doubled, we have:

y = 3x/2a + 5

A linear equation is represented as

y = mx + b

Where m represents the slope.

So, we have:

m1 = 3/a

m2 = 3/2a

Substitute m1 = 3/a in m2 = 3/2a

m2 = 1/2 * m1

Hence, the resulting equation will represent a line whose slope is 1/2 times the slope of the line

Read more about linear equation at:

brainly.com/question/14323743

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6 0
2 years ago
Where are the asymptotes for the following function located? f (x) = StartFraction 14 Over (x minus 5) (x + 1) EndFraction
m_a_m_a [10]

Answer:

2 vertical asymptotes occurring at x = 5 and x = -1

Step-by-step explanation:

given

f(x) = \frac{14}{(x-5)(x+1)}

recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)

Equating the denominator to zero,

(x-5)(x+1) = 0

(x-5) =0

x = 5 (first asymptote)

or (x+1) = 0

x = -1  (2nd asymptote)

6 0
3 years ago
) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle
Fynjy0 [20]

The surface (call it S) is a triangle with vertices at the points

x=0,y=0\implies z=2\implies(0,0,2)

x=0,y=2\implies z=-2\implies(0,2,-2)

x=2,y=0\implies z=-8\implies(2,0,-8)

Parameterize S by

\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to S to be

\vec s_v\times\vec s_u=(20v,8v,4v)

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv

=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}

8 0
3 years ago
18 times what is 504
schepotkina [342]

Answer:

28

Step-by-step explanation:

3 0
3 years ago
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