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tekilochka [14]

2 years ago

5

Explain why 6:4 and 18:8 are not equivalent ratios

2 answers:

Westkost [7]2 years ago

4 0

Answer:

False, 6:4 and 18:8 are equal ratios.

Step-by-step explanation:

Given data:

6:4 ≠ 18:8

Explanation.

1. 6:4 = 6/4 = 3/2

2. 18:8= 18/8 = 9/4 = 3/2

Hence proved:

6:4 = 18:8

Ludmilka [50]2 years ago

3 0

Answer:

Step-by-step explanation:

The given ratios are 6:4 and 18:8

This is not clear from the given ratios that they are equivalent.

We further simplify these ratios.

6 : 4 = $\frac{6}{4}=\frac{3}{2}$

18 : 8 = $\frac{18}{8}=\frac{9}{4}$

Since these fractions are not equal therefore, these ratios can't be equivalent.

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Delicious77 [7]

QUESTION 1

$4 {p}^{2} = 2 \times 2\times {p} \times p$

QUESTION 2

$39 {b}^{3} {c}^{2} = 3 \times 13 \times {b} \times b \times b\times {c} \times c$

QUESTION 3

$100 {x}^{3} y {z}^{2} = {2} \times 2 \times {5} \times 5 \times {x} \times x \times x\times y \times {z} \times z$

QUESTION 4

$66 {d}^{4} = 2 \times 3 \times 11 \times {d} \times d \times d \times d$

QUESTION 5

$85 {x}^{2} {y}^{2} = 5 \times 17 \times {x} \times x \times {y} \times y$

QUESTION 6

$18xy = 2 \times {3}^{2} \times x \times y$

$36 {y}^{2} = {2}^{2} \times {3}^{2} \times {y}^{2}$

The greatest common factor is the product of the least powers of the common factors;

$GCF=2 \times {3}^{2} \times y$

$GCF=18y$

QUESTION 7

$25mn = {5}^{2} \times m \times n$

$21m = 3 \times 7 \times m$

$GCF=m$

QUESTION 8

$15a = 3 \times 5 \times a$

$28a {b}^{2} = {2}^{2} \times 7 \times a \times {b}^{2}$

$GCF=a$

QUESTION 9

$24 {d}^{2} = {2}^{3} \times 3 \times {d}^{2}$

$30 {c}^{2} d = 2 \times 3 \times 5 \times d$

$GCF=2 \times 3 \times d$

$GCF=6d$

QUESTION 10

$20gh = {2}^{2} \times 5 \times g \times h$

$36 {g}^{2} {h}^{2} = {2}^{2} \times {3}^{2} \times {g}^{2} \times {h}^{2}$

$GCF= {2}^{2} \times g \times h$

$GCF=4gh$

QUESTION 11

$17 {c}^{3} {d}^{2} = 17 \times {c}^{3} \times {d}^{2}$

$5 {c}^{2} {d}^{2} = 5 \times {c}^{2} \times {d}^{2}$

$GCF= {c}^{2} \times {d}^{2}$

$GCF= {c}^{2}{d}^{2}$

4 0

2 years ago