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4 years ago

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How do you find a scale factor in Geometry

2 answers:

Bond [772]4 years ago

8 0

Answer: In two similar geometric figures, the ratio of their corresponding sides is called the scale factor. To find the scale factor, locate two corresponding sides, one on each figure. Write the ratio of one length to the other to find the scale factor from one figure to the other.

KonstantinChe [14]4 years ago

7 0

To find a scale factor between two similar figures, find two corresponding sides and write the ratio of the two sides. If you begin with the smaller figure, your scale will be less than one. If you start with the larger figure, your scale factor will be greater than one.

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Peyton buys cookies and apples at the store.

Dmitrij [34]

Answer:

Hope this helps sorry I'm in class right now

Step-by-step explanation:

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3 years ago

Write an equation to represent each function. Help with these two

nikklg [1K]

10y=x because if you multiply your y by 10, you would get the x

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4 years ago

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Prove that if the set of vectors {u1, u2, u3} is linearly dependent and u4 is any vector, then the set {u1, u2, u3, u4} is linea

Lesechka [4]

Answer with Step-by-step explanation:

We are given that the set of vectors ${u_1,u_2,u_3}$ is lineraly dependent set .

We have to prove that the set ${u_1,u_2,u_3,u_4}$ is linearly dependent .

Linearly dependent vectors : If the vectors $u_1,u_2.u_3,u_4$

are linearly dependent therefore the linear combination

$a_1u_1+a_2u_2+a_3u_3+a_4u_4=0$

Then ,there exit a scalar which is not equal to zero .

Let $a_1\neq0$ then the vector $u_1$ will be zero and remaining other vectors are not zero.

Proof:

When $u_1,u_2,u_3$ are linearly dependent vectors therefore, linear combination of vectors of given set

$a_1u_1+a_2u_2+a_3u_3=0$

By definition of linearly dependent vector

There exist a scalar which is not equal to zero.

Suppose $a_1\neq 0$ then $u_1=0$

The linear combination of the set ${u_1,u_2,u_3,u_4}$

$a_1u_1+a_2u_2+a_3u_3+a_4u_4=0$

When $a_1\neq0\; and\; u_1=0$

Therefore,the set ${u_1,u_2,u_3,u_4}$ is linearly dependent because it contain a vector which is zero.

Hence, proved .

5 0

3 years ago

Above is a table with missing terms. Come up with a possible common difference that makes sense and list the missing terms.

Rina8888 [55]

Answer:

common ratio:2 2. 18 3.36 4.72 equation: 4.5( $2^x$ )

Step-by-step explanation:

144=2*2*2*2*3*3 9=3*3

the common ratio would be 2

2. 9*2=18

3. 18*2=36

4. 36*2=72

(check) 72*2=144

0. 9/2=4.5

equation: 4.5( $2^x$ )

7 0

3 years ago

In the roof truss, line BG bisects ∠ABC and ∠DEF, m∠ABC= 112°, and ∠ABC≅∠DEF.

Dennis_Churaev [7]

In the given diagram, line BG bisects ∠ABC and ∠DEF, m∠ABC= 112°, and ∠ABC≅∠DEF. So the measure of angles are :

1. m∠DEF= 112° because ∠ABC≅∠DEF

2. m∠ABG= 56° because a straight line BG is bisecting ∠ABC in two equal parts. So, $\frac{112}{2} = 56$

3. m∠CBG= 56° because ∠ABG and ∠CBG are equal.

4. m∠DEG= 56° because ∠ABG ≅ ∠DEG

7 0

3 years ago

Read 2 more answers