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2 years ago

Which expressions are equivalent to 3(4h + 2k)?

Mathematics

1 answer:

faltersainse [42]2 years ago

5 0

Answer:

Step-by-step explanation:

If we distribute the 3 to the equation in the parenthesis, we end up with this

12h + 6k

Now, let's do the same thing for the other two equations

3(2k + 4h) = 6k + 12h

You can rearrange this equation to look like 12h+ 6k, making it the exact same as 3(4h + 2k).

So A is equivalent

How about B?

3(4h + 2h) = 12h + 6h = 18h

18h does not equal 12h + 6k, so B is not equivalent

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Answer:

20 over 36

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7 0

2 years ago

Read 2 more answers

Choose all the equations for which x=2 is a solution

saul85 [17]

Answer:

Option A and Option E

Step-by-step explanation:

Option A.

x + 3 = 5

(x + 3) - 3 = 5 - 3

x = 2

Option B.

x + 2 = 8

(x + 2) - 2 = 8 - 2

x = 6

Option C.

x + 1 = 1

(x + 1) - 1 = 1 - 1

x = 0

Option D.

x - 2 = 4

(x - 2) + 2 = 4 - 2

x = 2

Option E.

x - 7 = -5

(x - 7) + 7 = -5 + 7

x = 2

Therefore, Option (A) and Option (E) are the correct options.

5 0

3 years ago

What is the sum?<br> x2-9<br> X+3

VashaNatasha [74]

The sum is x2 - 9x + 3

4 0

3 years ago

Point I is on line segmentHJ‾HJ . Given HI=x,HI=x, IJ=2x+9,IJ=2x+9, and HJ=4x, HJ=4x, determine the numerical length of IJ‾.IJ .

Shalnov [3]

Answer:

IJ= 27

Step-by-step explanation:

HI+IJ=HJ

x+2x+9=4x

combine like terms

x=9

IJ=2(9)+9

18+9=27

IJ=27

4 0

3 years ago

The line l is tangent to the circle with equation x^2 + y^2=10 at the point P.

babunello [35]

Given:

The equation of a circle is

$x^2+y^2=10$

A tangent line l to the circle touches the circle at point P(1,3).

To find:

The equation of the line l.

Solution:

Slope formula: If a line passes through two points, then the slope of the line is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Endpoints of the radius are O(0,0) and P(1,3). So, the slope of radius is

$m_1=\dfrac{3-0}{1-0}$

$m_1=\dfrac{3}{1}$

$m=3$

We know that the radius of a circle is always perpendicular to the tangent at the point of tangency.

Product of slopes of two perpendicular lines is always -1.

Let the slope of tangent line l is m. Then, the product of slopes of line l and radius is -1.

$m\times m_1=-1$

$m\times 3=-1$

$m=-\dfrac{1}{3}$

The slope of line l is $-\dfrac{1}{3}$ and it passs through the point P(1,3). So, the equation of line l is

$y-y_1=m(x-x_1)$

$y-3=-\dfrac{1}{3}(x-1)$

$y-3=-\dfrac{1}{3}(x)+\dfrac{1}{3}$

Adding 3 on both sides, we get

$y=-\dfrac{1}{3}x+\dfrac{1}{3}+3$

$y=-\dfrac{1}{3}x+\dfrac{1+9}{3}$

$y=-\dfrac{1}{3}x+\dfrac{10}{3}$

Therefore, the equation of line l is $y=-\dfrac{1}{3}x+\dfrac{10}{3}$ .

4 0

2 years ago