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3 years ago

A rectangular drawing is enlarged by 30%. The original dimensions of this drawing are 16cm x 24cm.

Mathematics

1 answer:

Phoenix [80]3 years ago

4 0

Answer:

Step-by-step explanation: Scale $\frac{130}{100} = \frac{13}{10}$

New dimensions $16 * 1.3 --- 24*1.3 =20.8 cm * 31.2 cm$

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X^(log_(\sqrt(x))(x-3))=4

Zina [86]

Answer:

X= -4

Step-by-step explanation:

3 0

3 years ago

Solve 3/5=8/y. Round to the nearest tenth. 15.2 9.8 13.3 18.8

Sidana [21]

3/5 = 3/y
40 = 3y
Divide both by 3
Answer= 13.3

4 0

3 years ago

Identify rational numbers and intergers from the following

Nina [5.8K]

Answer:

Rational: all

Integers: 4, -36, -6

Step-by-step explanation:

A rational number is a number that can be written as a fraction of integers.

All numbers shown are rational numbers.

The integers are all positive and negative whole numbers and zero.

The integers are: 4, -36, -6

6 0

3 years ago

On a bicycle, Miranda rides for 4 hours and is 50 miles from her house. After riding for 6 hours, she is 74 miles away. What is

Vera_Pavlovna [14]

<h2> Miranda's rate is 12 hours per mies.</h2>

Step-by-step explanation:

Given by question,

Miranda began her ride from home merely that she was 50 miles away from home after 4 hours of riding.

To find, Miranda's rate = ?

2 hours later she was 974 miles from home.

∴ 2 hours she rode = 74 miles - 50 miles

Miranda rides 24 miles in 2 hours.

∴ Miranda rides in 1 hour = $\dfrac{24}{2}$ $\dfrac{miles}{hour}$

= $12$ $\dfrac{miles}{hour}$

Thus, Miranda's rate is 12 hours per mies.

3 0

3 years ago

Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco

Elza [17]

$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}$

$5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}$

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1 year ago