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ZanzabumX [31]
3 years ago
12

Apollo Spas services 156 hot tubs. If each hot tub needs 105 mL of muriatic acid, how many liters of acid are needed for all of

the hot tubs?
Mathematics
1 answer:
True [87]3 years ago
5 0

Answer:

16380 mL or 16.380 L

Step-by-step explanation:

1 hot tub requires 105 mL of Acid

156 hot tubs requires 105 * 156 mL of acid = 16380 mL of acid are needed.

16380 / 1000 mL/L = 16.380 Liters.

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The fifth grade teachers took their classes on a field trip to museum first group had 25 students and two teachers and cost $97.
iren [92.7K]

Answer:


Step-by-step explanation:

S=student cost; T=teacher cost

.

32S+3T=$127 . Multiply this by 2.

64S+6T=$254 Subtract this below

.

25S+2T=$97.50 . Multiply this by 3.

75S+6T=$292.50

64S+6T=$254 . Subtract this (from above.)

.

11S=$38.50

S=$3.50

ANSWER 1: Student tickets cost $3.50.


.

32S+3T=$127

32($3.50)+3T=$127

$112+3T=$127

3T=$15

T=$5.00

ANSWER 2: Teacher tickets cost $5.00.

.

CHECK:

25S+2T=$97.50

25($3.50)+2($5.00)=$97.50

$87.50+$10.00=$97.50

$97.50=$97.50


4 0
3 years ago
Solving systems equations by elimination
pishuonlain [190]
I won't answer them all for you (don't want to cheat - need to learn something).
But for the first one given:

<span>-2x-9y=-25
-4x-9y=-23

you set up either x or y (which ever one you want to eliminate) so that one of them cancels.
So this one is easy - you just have to multiply an entire row by negative (doesn't matter which one) - so that the y's will be + and - 

Since they're both 9 and they're subracting NOW - you can eliminate the y value. Now add/subtract the x's. So it'd be -2 or 2 depending on how you eliminated y. Then add the number outside of the = sign. Now just solve for x. After that plug in whatever you got for x into both of the rows. Then find the y value.

Hopefully this wasn't confusing -- reply if it is and I will try to simplify. I can give you contact if you need help with the rest 

Good luck!</span>
3 0
3 years ago
What assumption should be made to prove this conjecture by contradiction? If it is raining, then I will not go swimming.
Anon25 [30]
Here are some options to make deciding easier-

A) you went swimming
B) it is raining
C)you are not swimming
D) it is not raining


I think that the answer would be A because the opposite of it not raining would be it raining and the effect of that is that you will go swimming.

Hope this helps... plz mark Brainliest :)
8 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
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alexandr1967 [171]
Remember Rise/Run

So in this one you would start at -7 on the y-intercept, and then go down one, and over to the right two.
7 0
3 years ago
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