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3 years ago

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1 answer:

OlgaM077 [116]3 years ago

4 0

The answer is b...........

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What is the value of x?

Dafna1 [17]

Answer:

$3x = 12x - 81$ (vertically opposite angles)

$→3x - 12x = - 81$

$→- 9x = - 81$

$→x = \frac{ - 81}{ - 9}$

$→x = \frac{81}{9}$

$→x = 9$

hope it helped you:)

5 0

3 years ago

If there are 36 boys and 54 girls in a room, fill out all of the possible ratios of boys to girls that could be made.

aleksandr82 [10.1K]

Answer:

2:3

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Last year, Dale mowed the lawn in 45 min. This year, it took Dale 38 min to mow the lawn. What is the percent of decrease in mow

klasskru [66]

45-38= 7min decrease
(7/45)* 100=15.5%

3 0

3 years ago

Assume that a company sold 5.75 million motorcycles and 3.5 million cars in the year 2010. The growth in the sale of motorcycles

OverLord2011 [107]

Answer:

Final answer is approx 6.644 years.

Step-by-step explanation:

Given that a company sold 5.75 million motorcycles and 3.5 million cars in the year 2010. The growth in the sale of motorcycles is 16% every year and that of cars is 25% every year.

So we can use growth formula:

$A=P\left(1+r\right)^t$

Then we get equation for motorcycles and cars as:

$A=5.75\left(1+0.16\right)^t$

$A=3.5\left(1+0.25\right)^t$

Now we need to find about when the sale of cars will be more than the sale of motorcycles. So we get:

$3.5\left(1+0.25\right)^t>5.75\left(1+0.16\right)^t$

$3.5\left(1.25\right)^t>5.75\left(1.16\right)^t$

$3.5\left(1.25\right)^t>5.75\left(1.16\right)^t$

$\frac{\left(1.25\right)^t}{\left(1.16\right)^t}>\frac{5.75}{3.5}$

$\left(\frac{1.25}{1.16}\right)^t>1.64285714286$

$t\cdot\ln\left(\frac{1.25}{1.16}\right)>\ln\left(1.64285714286\right)$

$t>\frac{\ln\left(1.64285714286\right)}{\ln\left(\frac{1.25}{1.16}\right)}$

$t>6.6436473051$

Hence final answer is approx 6.644 years.

7 0

3 years ago

is <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3to%20%5C%3A%20the%20%5C%3A%20power%20%5C%3A%20of%20%5C%3A%20100%7D%7B3to%20%5

Norma-Jean [14]

Remember the laws of exponents:

$\frac{a^n}{a^m}=a^{n-m}$

Then:

$\begin{gathered} \frac{3^{100}}{3^{99}}=3^{100-99} \\ =3^1 \\ =3 \end{gathered}$

Therefore:

$\frac{3^{100}}{3^{99}}=3$

7 0

11 months ago