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4 years ago

15

Two parallel lines are crossed by a transversal.

1 answer:

kondaur [170]4 years ago

7 0

if I am right than right angle is equal to 90 degree

but here the answer or the value of my will be 78 degre cause opposite side of a parallels linse cut by transversal are equall

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Which of the following best describes the expression 5(x + 2)? The sum of constant factors 5 and x + 2 The product of constant f

aksik [14]

5 is a constant. (it has no variables)
x + 2 is not a constant. It does, however, have two terms, x and 2.
(terms are seperated by addition/subtraction)

This rules out a and b.

As for c... 5(x+2) implies <em>multiplication</em>, which would be a product, not a sum.

d. The product of a constant factor 5 and a 2-term factor x+2

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3 years ago

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HELP ASAP! WILL MARK BRAINLYNESS OR WHATEVER ITS CALLED

aksik [14]

Answer:

the first one.

Step-by-step explanation:

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3 years ago

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In a sale a gameboys are reduced by two fifths. what is the sale price if the original price was 50

vladimir1956 [14]

Answer:

The sale price would be $30.

Step-by-step explanation:

2/5 is like saying 20/50. If you take 50 and subtract 20 from that, you will be left with 30.

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3 years ago

What is the rate of growth, as a percent, for the following function? D(x)=23(1.0032)x

maks197457 [2]

I will assume you meant D(x)=23(1.0032)^x, if so:

The growth rate as a percentage is 100(.0032)=0.32% (about 1/3 of 1%)

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3 years ago

Point B has coordinates (4,1). The x-coordinate of point A is -4. The distance between point A and point B is 10 units. What a

klasskru [66]

Given:

Point B has coordinates (4,1).

The x-coordinate of point A is -4.

The distance between point A and point B is 10 units.

To find:

The possible coordinates of point A.

Solution:

Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between point A and point B is 10 units.

$\sqrt{(4-(-4))^2+(1-y)^2}=10$

Taking square on both sides, we get

$(8)^2+(1-y)^2=100$

$(1-y)^2=100-64$

$(1-y)^2=36$

Taking square root on both sides, we get

$(1-y)=\pm \sqrt{36}$

$-y=\pm 6-1$

$y=1\mp 6$

$y=1-6$ and $y=1+6$

$y=-5$ and $y=7$

Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).

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3 years ago