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Dmitrij [34]
3 years ago
5

Simplify 3 ∙ 2x. What is the coefficient?

Mathematics
2 answers:
matrenka [14]3 years ago
5 0
<span>3 ∙ 2x = 6x
so </span><span>coefficient = 6
</span>
hope it helps
Brilliant_brown [7]3 years ago
4 0
Hello,
simplify the expression:
3 · 2x = 6x
the coefficient is a number that stands right before x: kx
then in our case:
kx=6x
so k=6 is the coefficient

Bye :-)
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3 years ago
A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base
Olenka [21]

Answer:

The base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = x\times\frac{1}{2} =\frac{x}{2} \ m.

The height of triangle = x\times \frac{3}{4} =\frac{3x}{4}\ m.

The area of triangle = 16\ m^2.

Now, we put the formula of area to solve:

Area=\frac{1}{2} \times base\times height

16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}

16=\frac{3x^2}{16}

<em>Multiplying both sides by 16 we get:</em>

<em />256=3x^2<em />

<em>Dividing both sides by 3 we get:</em>

<em />\frac{256}{3} =x^2<em />

<em>Using square root on both sides we get:</em>

\frac{16}{\sqrt{3}}=x

x=\frac{16}{\sqrt{3}}

Now, by substituting the value of x to get the base and height:

Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.

<em>So, the base of triangle = </em>\frac{8}{\sqrt{3}} \ m.<em />

Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.

<em>Thus, the height of triangle =  </em>\frac{12}{\sqrt{3}} \ m.<em />

Therefore, the base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

7 0
3 years ago
Math help please TY :3<br><br> HWAJDSK 10 POINTS REWARDING !!
valkas [14]

I will do Point A carefully, The others I will indicate. Start with the Given Point A. Then do the translations

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Reflection: about x axis:x stays the same; y becomes -y:Result(-1,-2)

T<-3,4>: x goes three left, y goes 4 up (-1 - 3, -2 + 4): Result(-4,2)

R90 CCW: Point (x,y) becomes (-y , x ) So (-4,2) becomes(-2, - 4): Result (-2, - 4)

B(4,2) Original Point

  • Reflection: (4, - 2)
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C(4, -5) Original Point

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D(-1 , -5) Original Point

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Note: CCW means Counter Clockwise

The graph on the left is the same one you have been given.

The graph on the right is the same figure after all the transformations

7 0
3 years ago
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ycow [4]

Answer:

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Step-by-step explanation:

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3 years ago
Find the area of the region that lies under the parabola y=5x - x^2, where 1≤x≤4. I would definitely like to see some work :)
Elena L [17]

Answer:

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Step-by-step explanation:

You are expected to integrate the function between x=1 and x=4:

  \displaystyle\text{area}=\int_1^4{(5x-x^2)}\,dx=\left.\left(\dfrac{5}{2}x^2-\dfrac{1}{3}x^3\right)\right|_{x=1}^{x=4}\\\\=\dfrac{5(4^2-1^2)}{2}-\dfrac{4^3-1^3}{3}=37.5-21=\boxed{16.5}

__

<em>Additional comment</em>

If you're aware that the area inside a (symmetrical) parabola is 2/3 of the area of the enclosing rectangle, you can compute the desired area as follows.

The parabolic curve is 4-1 = 3 units wide between x=1 and x=4. It extends upward 2.25 units from y=4 to y=6.25, so the enclosing rectangle is 3×2.25 = 6.75 square units. 2/3 of that area is (2/3)(6.75) = 4.5 square units.

This region sits on top of a rectangle 3 units wide and 4 units high, so the total area under the parabolic curve is ...

  area = 4.5 +3×4 = 16.5 . . . square units

4 0
2 years ago
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