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3 years ago

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Ann and Joe’s father donated $3 for every lap they swam in a swim-a-thon. Ann swam 21 laps, and Joe swam 15 laps. Find the amoun

t of money their father donated.

1 answer:

DIA [1.3K]3 years ago

5 0

Step 1: Find the total number of laps that Ann and Joe did.

Ann: 21 laps
Joe: 15 laps
Total: 21 laps + 15 laps = 36 laps

Step 2: Multiply each lap swam by $3.

36 x $3 = $108

Their father donated $108.

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(x + y + 3)(x + y - 4)

ohaa [14]

Answer:

(X+Y+3)(X+Y-4)

= X^2+XY-4X+XY+Y^2_4Y+3X+3Y-12

=X^2+Y^2+2XY-X-Y-12

Step-by-step explanation:

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3 years ago

PLEASE HELP!! NEED ASAP 100 POINTS!!

Leto [7]

Answer:

See Below.

Step-by-step explanation:

Remember that the maximum account balance Ebony had was $400.

We can see that her bank balance first reached $400 on Day 4 and continued until Day 8.

Part A)

Since the maximum amount is $400, it is whenever the graph reaches (and stays) at its maximum, or the highest points on the graph.

We can see that at around Day 4, she first reached her maximum of $400 since that is the highest point of the graph.

And this continues until approximately Day 8.

Therefore, Ebony first reached $400 on Day 4 and stayed at $400 until Day 8.

Part B)

Refer above. The explanation is the same.

Part C)

Jade is says that the amount Ebony is depositing per day from Days 0 to 4 is the same as the amount Ebony is withdrawing per day from Days 8 to 12.

We also know that Day 12 is when Ebony’s account first reached 0.

Essentially, Jade is saying that the <em>slope</em> of the line from Days 0 to 4 is <em>the same</em> as the <em>slope</em> of the line from Days 8 to 12.

Ignoring the negative, we can immediately see that this is not true without calculating.

This is because the slope of the line from Days 8 to 12 is much steeper than than slope of the line from Days 0 to 4.

So, the two slopes are not equal since they are not the same steepness.

Therefore, this means that from Days 8 to 12, Ebony withdrew money at a <em>faster rate per day</em> than she had been depositing from Days 0 to 4.

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3 years ago

Read 2 more answers

The manager of a snack bar buys bottled water in packs of 35 and candy bars in packs of 20. Then, she sells the items individual

Lerok [7]

$\Huge\bf{\underline{\red \dag { Aиѕωєя }}} \red \dag$

4 Packs of water and 7 Packs of candy bars.

So, we wanna know the smallest number that both 35 and 20 will go into.

Find the Least Common Multiple (LCM), but to find the LCM we need to find the prime factorization of each of the following number :-

$\sf {35 = 5(7)}$

$\sf \red {20 = 5(4)}$

$\sf {4 = 2(2)}$

$\sf \red {20 = 5(2)(2)$

~Now multiply all the numbers by 5 :-

$\sf \pink{5(7)(2)(2) = 35(2)(2) = 70(2) = 140}$

This means she needs 140 bottles of water and 140 candybars.

Water is sold in packs of 35, this means that she needs :

$\sf \orange{ \frac{140}{35} = 4\: packs \: of \: water}$

Candy bars are sold in packs of 20, this means she needs :

$\sf \green{ \frac{140}{20} = 7 \: packs \: of \: candy \: bars}$

<u>___________________</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

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3 years ago

Given the function ƒ(x) = 20x2 − 18x − 25, find ƒ(−1).

dlinn [17]

f(x) = 20x² - 18x - 25

f(-1) = 20(-1)² - 18(-1) - 25

= 20 + 18 - 25

= 13

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3 years ago

una ventana rectangular tiene l metros de ancho y h metros de altura, con un perímetro de 6 metros y un área de 2m² ¿cuál es la

seropon [69]

Answer:

The formula for calculating the width of the window is

$w=\frac{-(-3)(+/-)\sqrt{-3^{2}-4(1)(2)}} {2(1)}$

Step-by-step explanation:

<u><em>The question in English is</em></u>

A rectangular window is l meters wide and h meters high, with a perimeter of 6 meters and an area of 2m². What is the formula for calculating the width of the window?

we know that

The perimeter of the window is equal to

$P=2(l+w)$

we have

$P=6\ m$

so

$6=2(l+w)$

simplify

$3=(l+w)$

isolate the variable l

$l=3-w$ ----> equation A

The area of the window is equal to

$A=lw$

we have

$A=2\ m^2$

so

$2=lw$ ----> equation B

substitute equation A in equation B

$2=(3-w)w$

solve for w

$2=3w-w^2$

$w^2-3w+2=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$w^2-3w+2=0$

so

$a=1\\b=-3\\c=2$

substitute in the formula

$w=\frac{-(-3)(+/-)\sqrt{-3^{2}-4(1)(2)}} {2(1)}$ ---> formula for calculating the width of the window

$w=\frac{3(+/-)\sqrt{1}} {2}$

$w=\frac{3(+/-)1} {2}$

$w=\frac{3(+)1} {2}=2\ m$

$w=\frac{3(-)1} {2}=1\ m$

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3 years ago