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RoseWind [281]
2 years ago
5

Please help with this question

Mathematics
1 answer:
NISA [10]2 years ago
5 0

Answer:

choice A

Step-by-step explanation:

f(x) * g(x) = √2x × √18x

= √36 * x

= 6x

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What is the equation of the line that passes through (-3, -1) and has a slope of 2/5 ? Put your answer in slope-intercept form.
Scrat [10]

Answer:

  a.  y = 2/5x + 1/5

Step-by-step explanation:

In point-slope form, the equation of a line with slope m through point (h, k) can be written

  y = m(x -h) +k

Then a line with slope 2/5 through the point (-3, -1) will have point-slope equation ...

  y = (2/5)(x +3) -1

This can be simplified to the desired form:

  y = 2/5x +6/5 -1 . . . . . . eliminate parentheses; next, collect terms

  y = 2/5x + 1/5

8 0
3 years ago
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How do i figure this out and get the correct answer?
Monica [59]
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FIND THE PERCENTAGE OF THE PART!!!!!!<br> 15% of 30 of what number???
lawyer [7]

Answer:

4.5

Step-by-step explanation:

15% of 30 is 4.5

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2 years ago
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1. Name the figure (It is not a cube; it is not a cylinder; what is it?)
Arlecino [84]
1. It is a triangular prism

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3 years ago
A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co
butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

\displaystyle \vec{v}=\frac{\vec{x}}{t}

Where \vec x is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

\displaystyle \vec{x}=\vec{v}.t

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=\ km/h

The displacement of the plane in 2 hours is

\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)

\displaystyle \vec{x_1}=km

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=km/h

The displacement in 1 hour is

\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)

\displaystyle \vec{x_2}=km

The total displacement is the vector sum of both

\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+

\displaystyle \vec{x_t}=km

\displaystyle \vec{x_t}=

The distance from the airport is the module of the displacement:

\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}

\displaystyle |\vec{x_t}|=894.43\ km

8 0
3 years ago
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