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svetlana [45]
3 years ago
11

G(x) = -10x – 8 a function

Mathematics
1 answer:
wlad13 [49]3 years ago
7 0

Answer:

yes that is a function and not an equation

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carries family leaves at 7:15. they drive for 30 minutes and yhen stop for dinner what time is it when they stop for dinner?
laiz [17]

Answer:

I think it is 7:45

Step-by-step explanation:


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Identify the property that the statement illustrates
oksian1 [2.3K]

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6 units away from 0.

Step-by-step explanation:

when move to 6 you count the units.

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Which ratio can you get by scaling up the ratio 5/6 / 1/3? Select all that apply.
Liula [17]

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the answer is 5/2 or alternative form 2*1/2 so the answer is C

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Solve the solution of the equations 10x-3y=3 and 2x+y=9
forsale [732]

Answer:

x=1.875   y=-0.875

Step-by-step explanation:

10x-3y=3                     10x-3y=3

3(2x+y)=(9)3                +<u>6x+3y=27</u>

                               1/16(16x)=(30)1/16  = x=1.875

   10x-3y=3                  10x-3y=3

 -5(2x+y)=(9)-5           -<u>10x-5y=4  </u>                        

                                    1/-8(-8y)=(7)1/-8 = y=-0.875

4 0
3 years ago
Polly stacked the tins from five boxes of cat food onto an empty shelf in a supermarket. There were 15 tins of cat food in each
Alexandra [31]

Answer:

(a) 3\frac{2}{3} boxes of cat food had been sold.

(b) \frac{11}{3} boxes of cat food had been sold.

Step-by-step explanation:

From the question,

Polly stacked the tins from five boxes of cat food onto an empty shelf and there were 15 tins of cat food in each box. That is,

5 × 15 tins of cat food were stacked onto the empty shelf.

5 × 15 = 75

Hence, 75 tins of cat food were stacked onto the empty shelf.

Also, from the question,

At the end of the week there were 20 tins left on the shelf and the rest of the tins had been sold, that is

75 - 20 tins had been sold

75 - 20 = 55

Hence, 55 tins of cat food had been sold.

To determine the number of boxes of cat food that had been sold,

Since there are 15 tins of cat food per box, then

we will divide 55 tins of cat food that had been sold by 15 tins of cat food per box

55 ÷ 15 = \frac{55}{15}

Hence, \frac{55}{15} boxes of cat food had been sold

(a) As a mixed number

\frac{55}{15}  = 3\frac{2}{3}

Hence, 3\frac{2}{3} boxes of cat food had been sold.

(b) As an improper fraction

\frac{55}{15}  = \frac{11}{3}

Hence, \frac{11}{3} boxes of cat food had been sold.

7 0
3 years ago
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