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3 years ago

• If the new image is smaller than the old image would

Mathematics

1 answer:

mars1129 [50]3 years ago

5 0

Answer:

Less than one

Step-by-step explanation:

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What is the answer to 5 over 35 simplalifying

Finger [1]

Answer:

1/7

Step-by-step explanation:

Divide each number by 5.

5/5=1

35/5=7

8 0

3 years ago

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The work of a student to solve a set of equations is shown below: m=8+2n; 4m=4+4n. Solve by elimination

nadya68 [22]

Answer:

Step-by-step explanation:

4(8+2n) = 4 + 4n

32 + 8n = 4+4n

4n = -28

n = -7

4 0

3 years ago

Fill in the blanks thanks

Mademuasel [1]

Answer:

Um threre is no blanks or there is no question.

Step-by-step explanation:

5 0

3 years ago

Variable y varies directly with variable x, and y = 6 when x = 9.

lesantik [10]

X and y are proportional hence when y is multiplied by a number, x is multiplied by the same number.

18=6*3 hence when y=18,x=9*3=27

x=27

5 0

3 years ago

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Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$

6 0

3 years ago

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