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Shtirlitz [24]
3 years ago
14

Multiply by applying the commutative and/or associative property.

Mathematics
2 answers:
arlik [135]3 years ago
6 0
3.20 is it I will be w
vovangra [49]3 years ago
6 0

Answer:

maybe 126

Step-by-step explanation

did my best hope it helped.

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What is the equation of the line in slope-intercept form? plz help 20 points
AfilCa [17]

To find slope intercept form first do rise over run

rise 4 run 3

then find where it crosses the x-intercept to get

y=4/3x-3

5 0
4 years ago
The distance between City A and City B is 500 miles. A length of 1.8 feet represents this distance on a certain wall map. City C
Ratling [72]

Answer: 800 miles

Step-by-step explanation:

Let the actual distance between City C and City​ D be represented by x.

The information given in the question can be formed into an equation as:

1.8 feet / 2.88 feet = 500 miles / x

Cross multiply

1.8 × x = 500 × 2.88

1.8x = 1440

x = 1440/1.8

x = 800

Therefore, the actual distance between City C and City​ D is 800 miles.

3 0
3 years ago
Prime factorization of 52​
lara [203]

Answer:

The first prime factor we test is 2:

52 / 2 = 26

26 / 2 = 13

13 is a prime number so the prime factorization is

2 * 2 * 13 = 52

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Round 67,436,828,104 to the nearest million
S_A_V [24]
It would be 67,437,000,000 since the eight in 67,436,(8)28,104 rounds the 6 up to a 7.
4 0
3 years ago
Read 2 more answers
The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
maxonik [38]

Given:

The red figure dilated with a scale factor of s=\dfrac{1}{3} and the center of dilation is at the point (4,2) to get the green figure.

To find:

The coordinates of C' and A.

Solution:

If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

In given problem, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

4 0
3 years ago
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