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Cloud [144]
3 years ago
13

1/3y+1/4=5/12 i need 20 character

Mathematics
1 answer:
lys-0071 [83]3 years ago
3 0
Y/3 + 1/4 = 5/12
4y/12 + 3/12 = 5/12
           4y/12 = 2/12
                4y = 2
                /4     /4
                  y = 1/2

Therefore y = 1/2

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Find the missing value to the nearest hundredth. sin ___ = 7/12 (1 point) 35.69° 30.26° 80.16° 54.31°
Alisiya [41]
In your calculator, input arcsin(7/12). Make sure that your calculator is in degree mode. The answer is 35.69 degrees.
3 0
3 years ago
1. If 5x + 6 = 10, what is the value of 10x + 3?
Digiron [165]
Here the question is simple.
All because, we only need to find the value of x.
We are given two equations.
5x + 6 = 10 and 10x + 3 =?
So, we will find the the value of x in the first equation, so that we can substitute the value of x in the second one and there we are with the answer.
5x + 6 = 10
For finding the value of x, all we have to do is,
Transpose the number 6 to 10
Therefore. 5x = 10 - 6 ( Take the equal sign as The Magic Bridge on which if anyone crosses it , will change its sign.)
So we have,
5x = 4
So x = 4/5 ( Multiplication will change to division after crossing the equal sign)
( Doubtful? Substitute the value of x and try!)
Now that we got the value of x,
We can just simply substitute the value of x in the second equation.
10x + 3 = ?
x = 4/5
10*4/5 +3 => 5 and 10 get canceled to 2 at the numerator.
By normal multiplication and then addition, we will get,
8 + 3 = 11
Hope this helps!!!! :)
6 0
3 years ago
Read 2 more answers
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
An unmanned submarine ascends from the ocean floor at a rate of 2 meters per second. If the ocean floor is 4220 feet below sea l
KengaRu [80]

Answer:

a=4220-2t

This would be if we are finding the altitude of the submarine from the surface of the water(how many meters away from the surface it is). I don't know what it would be if you want to know from the bottom of the ocean.

4 0
3 years ago
Read 2 more answers
Three teachers handed out math and science textbooks for their classes two teachers had 21 students each and the last teacher ha
Tanya [424]

Answer:

64 textbooks

Step-by-step explanation:

the first two teachers had 21 students each that's 21+21=42

the other teacher had 22 students so apparently all the textbooks handed to the students are,21+21+22=64

8 0
2 years ago
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