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3 years ago

Tripping ran 4.8 times as many laps as Tony.If Tony ran3.7 laps how many laps did tripping run

Mathematics

2 answers:

Usimov [2.4K]3 years ago

7 0

This is a simple multiplication question.

4.8 x 3.7=17.76

Tripping ran 17.76 laps.

Check: 17/76/4.8=3.7

Fittoniya [83]3 years ago

5 0

You simply multiply 4.8 and 3.7 because if you notice the word "times," it explains you that you need to do multiplication.

4.8*3.7 = 17.76

So, Tripping ran 17.76 laps.

HOPE THIS HELPS!!!!

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Leslie and Daniel plan to put an above-ground pool in the backyard of their new home. The backyard is rectangular, 30 feet by 50

Brut [27]

1047.84ft² is not covered by the pool.
Find the area of the yard covered by pull using the area of a circle formula (the height is irrelevant in this case). If the diameter of the pool is 24 feet, its radius is 12 (half of the diameter)

A = 3.14r^2
A =3.14(144)
A = 452.16 ft²

Subtract the area of the pool from the area of the yard to get the area of the yard that is not covered by the pool. If the dimensions of the yard are 30ft by 50ft, you multiply them to get the area: 1500ft²

Total yard area: 1,500ft²
Area of yard without pool: 1,500ft² - 452.16ft² = 1047.84ft²

4 0

3 years ago

In a competitive examination negative marks of 2 are awarded for each wrong answers and 4 marks for every correct answer. If nee

Alex777 [14]

9514 1404 393

Answer:

Rajan gets 6 marks more

Step-by-step explanation:

Neerav's marks are (15)(4) +(5)(-2) = 60 -10 = 50.

Rajan's marks are (14)(4) = 56.

Rajan gets 6 more marks than Neerav.

3 0

2 years ago

Leo's bank balances at the end of months 1, 2, and 3 are $1500, $1530, and $1560.60,

grandymaker [24]

Leo's balance after 9 months will be: $1757.49

Step-by-step explanation:

It is given that the balances follow a geometric sequence

First of all, we have to find the common ratio

Here

$a_1 = 1500\\a_2 = 1530\\a_3 = 1560.60$

Common ratio is:

$r = \frac{a_2}{a_1} = \frac{1530}{1500} = 1.02\\r = \frac{a_3}{a_2} = \frac{1560.60}{1530} = 1.02$

So r = 1.02

The general form for geometric sequence is:

$a_n = a_1r^{n-1}$

Putting the first term and r

$a_n = 1500 . (1.02)^{n-1}$

To find the 9th month's balance

Putting n=9

$a_9 = 1500 . (1.02)^{9-1}\\= 1500.(1.02)^8\\=1757.4890$

Rounding off to nearest hundredth

$1757.49

Hence,

Leo's balance after 9 months will be: $1757.49

Keywords: Geometric sequence, balance

Learn more about geometric sequence at:

#LearnwithBrainly

5 0

3 years ago

the slide at the playground has a height of 6 feet.The base of the slide measured on the ground is 8 feet.what is the length of

NemiM [27]

The formula for finding the length is
${a}^{2} + {b}^{2} = {c}^{2}$
All we have to do is plug in the numbers.

${6}^{2} + {8}^{2} = {c}^{2}$
36 + 64 = 100

Since 100 =
${c}^{2}$
we have to find the square root.

$\sqrt{100} = 10$

The length of the slide is 10 feet.

6 0

3 years ago

For what value of a should you solve the system of elimination?

SIZIF [17.4K]

$\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}$

$\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20$
$\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12$

$\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}$

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

$\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}$

$3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$
$3a-10y-3a=-8-3a$

$\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10$
$\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}$

Simplify more.

$\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10}$

$\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y$

$-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}$

$\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}$

$\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20$

$10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}$
$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}$

$6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}$
$6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}$

$\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}$

$\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x$

$\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}$

$20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \ \frac{20}{1}-\frac{80}{-3a+10}$

$\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10$

$Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
$\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \ \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20$

$\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \ 20\left(-3a+10-4\right) \ \textgreater \ Factor\;more$

$10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite$
$-3a+2\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine$
$60\left(-a+2\right)$

$\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \ \frac{10\left(-a+2\right)}{\left(-3a+10\right)}$

$\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ \frac{10\left(-a+2\right)}{-3a+10}$

$Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}$

Hope this helps!

7 0

3 years ago

Read 2 more answers