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gogolik [260]
3 years ago
10

Is this right, d 3 cubic inits

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
4 0
Volume=length×width×<span>height
   = 1 X 1 X 1 =  1
YOUR ANSWE IS C THE ANSWER IS 1 CUBIC UNIT

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I will give BRAINLIEST to whoever is CORRECT
8090 [49]
I believe the answer could be 

h = -16t^2 + 32t + 12
      ---------------------------
                    t<span />
7 0
4 years ago
Need pre-cal help. Will mark best answer brainliest
OlgaM077 [116]

so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.


\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.


\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

5 0
3 years ago
195÷1.5=(195∙10)÷(1.5∙10)=
solong [7]

Answer:

130

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
3 = –(–y + 6)<br> a. –9<br> b. –3<br> c. 3<br> d. 9
Vlad1618 [11]
3 = -(-y + 6)

First, simplify brackets. / Your problem should look like: 3 = y - 6
Second, add 6 to both sides. / Your problem should look like: 3 + 6 = y
Third, simplify 3 + 6 to 9. / Your problem should look like: 9 = y
Fourth, switch sides. / Your problem should look like: y = 9

The answer is D) 9.

7 0
4 years ago
Find expressions for the partial derivatives of the following functions:
nlexa [21]

Answer:

Step-by-step explanation: partial derivative is the differentiation of one variable e.g. X while leaving the values of the other variable e.g. Y

These four questions A, B, C and D have different functions separated by commas. I will not assume the commas to be something else like a plus sign.

A. f(x) = g'(x).k(y) , g'(x) + h(y)

f(y) = k'(y).g(x) , g(x) + h'(y)

B. f(x) = g'x (x+y)

f(y) = g'y (x+y) , h'y (y+z)

f(z) = h'z (y+z)

C. f(x) = f'x (xy) , f'x (zx)

f(y) = f'y (xy) , f'y (yz)

f(z) = f'z (yz) , f'z (zx)

D. f(x) = f'x (x) , g'(x) , h'x (x,y)

f(y) = h'x (x,y)

These are the partial derivative expressions for each variable in each function. You will need to pay a lot of attention to understand:

* while differentiating X alone, functions in Y which are separated by commas from the functions in X, are ignored totally because they are different questions

* In functions where X added to Y is in a bracket e.g. (x+y), to find the derivative of X, Y isn't thrown away because they are joined (by a plus sign) the derivative of X alone in this case would be f'x (x+y)

* f(x), just like g(x), simply means/represents a function in X hence f'(x) means the differentiation of all X-terms in that function

6 0
3 years ago
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