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VladimirAG [237]
3 years ago
11

Anybody know this????

Mathematics
1 answer:
Dimas [21]3 years ago
4 0
I believe the answer Is 35. I remember there bing some rule about it. I hope this helps and I’m sorry if I confused you my brain is a little fried.
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A chemist is using 378 milliliters of a solution of acid and water. If 18.2% of the solution is acid, how many milliliters of ac
Wewaii [24]

Answer:

If 18.2% of the 378 milliliters of solution is acid, 0.182(378) ml of the solution is acid. Multiplying this out, our final answer is 68.8 ml of the solution is acid.

Let me know if this helps!

7 0
3 years ago
Read 2 more answers
Solve the following equation for h.<br> r+3Q/h=t
Arada [10]

Answer:

             \bold{h=\dfrac{3Q}{t-r}}

Step-by-step explanation:

\bold{r+\frac{3Q}h=t}\\-r\qquad\ -r\\\bold{\frac{3Q}h=t-r}\\\cdot h\qquad \cdot h\\ \bold{3Q=(t-r)h}\\^{\div(t-r)\quad\div(t-r)}\\\bold{\dfrac{3Q}{t-r}=h}

Or, if you mean (r+3Q)/h=t:

\bold{\frac{r+3Q}h=t}\\{}\ \ \cdot h\quad\ \cdot h\\\bold{r+3Q=ht}\\{}\ \ \div t\qquad \div t\\\bold{\dfrac{r+3Q}{t}=h}\\\\\bold{h=\dfrac{r+3Q}{t}}

4 0
3 years ago
Has one negative and one positive coefficient Simplifies to 5x + 12​
rosijanka [135]

Answer:

Step-by-step explanation:

whats the question here lol

8 0
2 years ago
Can someone answer part B for me? Thank you and will give brainliest.
frez [133]

Answer:

The answer to part B is B.

Step-by-step explanation:

Could you take another photo so I can see the bottom of the page?

4 0
3 years ago
Which equation best represents this situation?
Harlamova29_29 [7]
Hello!

We have the difference between 30 and x(our unknown value) is equal to 17; so we just simply set it up that way. Always think of x or any variable(like in your question they use p, it can be anything u want, it’s simply a representation of an unknown value we are trying to find)as your unknown and what operation they are implying as in this case it’s subtraction as they want the difference between 30 and x.

So the answer is a. 30-p=17

Hope this helped and any questions please just ask. Thank you!
7 0
3 years ago
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