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Zina [86]
3 years ago
15

Solve the equation for y: 6x-2y=14 Must show work

Mathematics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

y = -7 + 3x

Step-by-step explanation:

6x - 2y = 14

- 2y = 14 - 6x

y =  - 7 +  3x

y =  - 7 + 3x

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After substituting, what is the first step when evaluating 4y+9/7 when y=8
kenny6666 [7]

Answer:

  multiply 4 and 8 to get 32

Step-by-step explanation:

After substituting, the expression is ...

  4·8 +9/7

The order of operations tells you to do multiplication and division before addition and subtraction. You do them left-to-right. The multiplication on the left is 4·8, so you do that first. The result is ...

  32 + 9/7

Now, you can do the division:

  32 + (1 2/7)

And, finally, the addition:

  33 2/7

___

Or, you could skip the division and go straight to adding a whole number and a fraction:

  (32·7 +9)/7 = (224+9)/7 = 233/7

7 0
3 years ago
What's 950 dividend by 48
stepladder [879]
19.79 should be the correct answer?

4 0
3 years ago
Read 2 more answers
How to show the tens fact you used. write the difference. <br> 12-7=____<br> 10-___=____
prisoha [69]
The first answer is 5,but i don't know about the second one. (sorry)
3 0
3 years ago
The length of one side of a triangular lot is 6 m less than 3 times the length of the second side. The third side is 8 m longer
Tanya [424]

second side = x

first side = 3x - 6

third side = (3x - 6) + 8

The perimeter = 80m and is equal x + (3x - 6) + [(3x - 6) + 8]

x + 3x - 6 + 3x - 6 + 8 = 80

7x - 4 = 80 |+4

7x = 84 |:7

x = 12 m

3x - 6 = 3(12) - 6 = 30 m

(3x - 6) + 8 = 3x + 2 = 3(12) + 2 = 38 m

Answer: 12m, 30m, 38m

8 0
3 years ago
Express the given integral as the limit of a riemann sum but do not evaluate: the integral from 0 to 3 of the quantity x cubed m
WARRIOR [948]
We will use the right Riemann sum. We can break this integral in two parts.
\int_{0}^{3} (x^3-6x) dx=\int_{0}^{3} x^3 dx-6\int_{0}^{3} x dx
We take the interval and we divide it n times:
\Delta x=\frac{b-a}{n}=\frac{3}{n}
The area of the i-th rectangle in the right Riemann sum is:
A_i=\Delta xf(a+i\Delta x)=\Delta x f(i\Delta x)
For the first part of our integral we have:
A_i=\Delta x(i\Delta x)^3=(\Delta x)^4 i^3
For the second part we have:
A_i=-6\Delta x(i\Delta x)=-6(\Delta x)^2i
We can now put it all together:
\sum_{i=1}^{i=n} [(\Delta x)^4 i^3-6(\Delta x)^2i]\\\sum_{i=1}^{i=n}[ (\frac{3}{n})^4 i^3-6(\frac{3}{n})^2i]\\&#10;\sum_{i=1}^{i=n}(\frac{3}{n})^2i[(\frac{3}{n})^2 i^2-6]
We can also write n-th partial sum:
S_n=(\frac{3}{n})^4\cdot \frac{(n^2+n)^2}{4} -6(\frac{3}{n})^2\cdot \frac{n^2+n}{2}

4 0
3 years ago
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