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cestrela7 [59]
3 years ago
6

−3(y+5)=15 solve this math problem

Mathematics
2 answers:
Serhud [2]3 years ago
7 0

Answer:

y = -10

Step-by-step explanation:

Hey there!

To solve this you first have to distribute the -3 to y and 5. When you do this you get...

-3y - 15 = 15

Add 15 on both sides you get...

-3y = 30

Divide by -3 and you get...

y = -10

Hope this helped :)

Aleksandr [31]3 years ago
6 0

Answer:

y=-10

Step-by-step explanation:

Divide both sides by -3:

-3(y+5)/-3=15/-3

Simplify: 15/-3

y+5=-5

subtract 5 from both sides:

y+5-5=-5-5

simplify:

y=-10

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2x-6 (2) is your equation. follow these steps, and you have to distribute 6 into 2 like this: 2x-12. you get negative 12 because -6 x 2 =-12. now you divide -12 by 2 which is -6. So -6=x
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Which of the following relationships represents a function?
belka [17]

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3 years ago
3 2/3 divided by 3 4/5 <br> if you could can you give it to me now I'm in a hurry
soldier1979 [14.2K]

Answer:

<em>3 2/3 divided by 3 4/5 is</em> \frac{55}{57}

Step-by-step explanation:

<u>Operations With Fractions</u>

Let's divide 3 2/3 by 3 4/5. Both fractions are in mixed format, so we have to write them as an improper fraction:

\displaystyle 3\frac{2}{3}=3+\frac{2}{3}=\frac{9+2}{3}=\frac{11}{3}

\displaystyle 3\frac{4}{5}=3+\frac{4}{5}=\frac{15+4}{5}=\frac{19}{5}

We must divide \frac{11}{3} by \frac{19}{5}

To make it simple, we just multiply the first fraction by the reciprocal of the second fraction:

\displaystyle \frac{\frac{11}{3} }{ \frac{19}{5}}=\frac{11}{3} * \frac{5}{19}=\frac{55}{57}

3 2/3 divided by 3 4/5 is \frac{55}{57}

4 0
3 years ago
3. Could a 5 x 8 matrix have dim Col A 6 and dim Nul A - 2? Justify your answer. (Hint: Think about the possible pivot rows and
alexdok [17]

Answer:

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Step-by-step explanation:

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4 years ago
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Answer: h(x)=10(0.4)^x

Step-by-step explanation:

The exponential function h, represented in the table, can be written as h(x)=ab^x

From table, at x=0, h(x) =10

Put theses values in equation,, we get

10=a.b^0\\\\\Rightarrow\ 10=  a (1)\\\\\Rightarrow\ a= 10

Also, for x= 1 , h(x) = 4, so put these values and a=10 in the equation , we get

4=10b^1\\\\\Rightarrow\ b=\dfrac{4}{10}\\\\\Rightarrow\ b= 0.4

Put value of a and b in the equation ,

h(x)=10(0.4)^x  →  Required equation.

5 0
3 years ago
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