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2 years ago

Determine whether the statement below is true or false boxy is rectangle

Mathematics

1 answer:

Flauer [41]2 years ago

8 0

True I think I am not Fr Fr tho

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What does x equal in -7=-1+x/3

Scorpion4ik [409]

Hello! :)

To solve your equation, we will first simplify both sides of the equation

-7 = -1 + x/3

-7 = -1 + 1/3x

-7 = 1/3x - 1

Next, we will flip the equation

1/3x - 1 = -7

Third, we will add 1 to both sides

1/3x - 1 + 1 = -7 + 1

1/3x = -6

Last, we will Multiply both sides by 3

3 * (1/3x) = (3) * (-6)

x = -18 (ANSWER)

That means -18 is you answer

Hope this helped you!

THEDIPER

3 0

3 years ago

You invest $5000 at 6.25% interest. How much is in your account after 8 years if

romanna [79]

Answer:

A. annually I believe it's the answer bc you add interest every year that way

6 0

3 years ago

Read 2 more answers

Write an algebraic expression for the phrase. The quotient of 5 and the sum of 12 and a number x

anzhelika [568]

Answer:

the answer is (12 +x)/5 please mark brainliest

3 0

2 years ago

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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Which equation has no solution?<br> |-x - 3| = 5<br> |2x – 1| =0<br> |15 - 3x|= -8<br> |-x + 9| = 0

djverab [1.8K]

Answer:

the last one

Step-by-step explanation:

5 0

3 years ago