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yawa3891 [41]
3 years ago
7

Which unit would you use to measure the amount of water in a bathtub?

Mathematics
1 answer:
zepelin [54]3 years ago
7 0
Liters the rest are to small and kg is for solids
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How do I solve the system of equations<br> {Y=x^2+2x}<br> {Y=-x-2}
Minchanka [31]

Graph it on the Ti-84 and use the trace function to find the intersection.

The answer is

(-3,3)

7 0
3 years ago
Renting a room at the Willow tree resort costs $125 for 2 people. Each additional person is $25. Write an expression that shows
Mama L [17]

Answer:

Let C be the cost for the people staying

if n <= 2 ;  C = $125

If n >2 ;  C =  $125 + $25 * (n-2)

Step-by-step explanation:

If the number of people is more than two, we need to add $25 for each of them.

Since n will be the number of people staying, we need to make sure that we substract two from this term to avoid overcharging the guests when there is less than two people.

So

(n-2) is the term that should be multiplied by a factor of 25.


5 0
3 years ago
Please help!! (no links)
attashe74 [19]

Answer:

3/32

Step-by-step explanation:

The first question should be 3/8 as there are 3 "1"s on the spinner. The answer to the last one is "3/32". You take the 3/8 chances to get a one and multiply that by 1/4 to get a "B". Once you multiply you get .09375 which is 3/32 when converted. Please let me know if this is incorrect.

5 0
2 years ago
HELP ASAPPPP PLEASE I NEED TO FINISH ASAP
defon

Answer:

-15.3, -15.1, -12.3, -11.4, -11.4, 19.6

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
If sin(theta) =4/5 and is in quadrant 2, the value of cot(theta
iren [92.7K]
\sin\theta= \frac{4}{5} &#10;\\&#10;\\\sin^2\theta+\cos^2\theta=1&#10;\\ \cos\theta=\pm \sqrt{1-\sin^2\theta} =\pm \sqrt{1-( \frac{4}{5})^2 } =\pm \sqrt{1- \frac{16}{25} } =\pm \sqrt{\frac{9}{25} }=\pm \frac{3}{5}}

\theta \in II \Rightarrow \cos\theta\ \textless \ 0 \Rightarrow \cos\theta=-\frac{3}{5}

\cot\theta= \frac{\cos\theta}{\sin\theta}= \frac{-\frac{3}{5}}{\frac{4}{5}}  =- \frac{3}{4}

5 0
3 years ago
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