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Sauron [17]
3 years ago
9

6:57 47 minutes before

Mathematics
2 answers:
Verdich [7]3 years ago
6 0
It would be 6:10 that should be it
ziro4ka [17]3 years ago
4 0
6:57 + 47 = 7: 44
or 6:10
................
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A piece of string 25 1/2 inch long will be cut into 3/4 inches pieces how many pieces will there be
weqwewe [10]
25.5/0.75 = 34 Pieces.

Hope this has helped.
6 0
3 years ago
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a)express x^2+12x+11 in the form (x+p)^2+q b)A curve has equation y=x^2+12x+11. What are the co-ordinates of the turning point o
sveticcg [70]

a) (x+6)^2-25

b) (-6, -25)

Step-by-step explanation:

Perfect square trinomial! If you have 12x, divide by two and get 6, so we have (x+6)^2+q. Then solve for q. You get q as -25.

For b, we just take the vertex from vertex form. We get x+6=0 so x=-6, and q = -25, so the answer is (-6, -25).

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4 0
3 years ago
I need helppp<br> It’s for math and I’m clueless
Tcecarenko [31]

Answer:

14

Step-by-step explanation:

x=4 y=1

Substitute them into the equation 3x-4yy+6

3(4)-4(1)+6

12-4+6

5 0
2 years ago
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Martha has 8 cubic feet ofputting soil in the three flowerpots she wants to put the same amount of soil in each pot how many cub
frozen [14]

Answer:

2\frac{2}{3}\text{ ft}^3\approx 2.67\text{ ft}^3

Step-by-step explanation:

We have been given that Martha has 8 cubic feet of putting soil in the 3 flowerpots. She wants to put the same amount of soil in each pot.

To find the amount of soil in each pot, we need to divide total soil (8 cubic feet) by number of pots (3) as shown below:

\text{Amount of soil in each pot}=\frac{8\text{ ft}^3}{3}

\text{Amount of soil in each pot}=2\frac{2}{3}\text{ ft}^3

\text{Amount of soil in each pot}=2.6666\text{ ft}^3

\text{Amount of soil in each pot}\approx 2.67\text{ ft}^3

Therefore, Martha needs to put approximately 2.67 cubic feet of soil in each flower pot.

8 0
2 years ago
Given f(x) = 6(1-X), what is the value of f(8)?
andrey2020 [161]

Answer:

-42

Step-by-step explanation:

f(8)=6(1-8)

because of that 1-8 is -7 so you times by 6 and get -42

8 0
2 years ago
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