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Natalka [10]
3 years ago
8

I need help solving these problems​

Mathematics
1 answer:
Delicious77 [7]3 years ago
4 0

QUESTION 1

4 {p}^{2}  =  2 \times 2\times  {p} \times p

QUESTION 2

39 {b}^{3}  {c}^{2}  = 3 \times 13 \times  {b} \times b \times b\times  {c} \times c

QUESTION 3

100 {x}^{3} y {z}^{2}  =  {2} \times 2 \times  {5} \times 5  \times  {x} \times x \times x\times y \times  {z} \times z

QUESTION 4

66 {d}^{4}  = 2 \times 3 \times 11 \times  {d} \times d \times d \times d

QUESTION 5

85 {x}^{2}  {y}^{2}  = 5 \times 17 \times  {x} \times x   \times {y} \times y

QUESTION 6

18xy = 2 \times  {3}^{2}  \times x \times y

36 {y}^{2}  =  {2}^{2} \times  {3}^{2}   \times  {y}^{2}

The greatest common factor is the product of the least powers of the common factors;

GCF=2 \times  {3}^{2}  \times y

GCF=18y

QUESTION 7

25mn =  {5}^{2}  \times m \times n

21m = 3 \times 7 \times m

GCF=m

QUESTION 8

15a = 3 \times 5 \times a

28a {b}^{2}  =  {2}^{2}  \times 7 \times a \times  {b}^{2}

GCF=a

QUESTION 9

24 {d}^{2}  =  {2}^{3}  \times 3 \times  {d}^{2}

30 {c}^{2} d = 2 \times 3 \times 5 \times d

GCF=2 \times 3 \times d

GCF=6d

QUESTION 10

20gh =  {2}^{2}  \times 5 \times g \times h

36 {g}^{2}  {h}^{2}  =  {2}^{2}  \times  {3}^{2}  \times  {g}^{2}  \times  {h}^{2}

GCF= {2}^{2}  \times g \times h

GCF=4gh

QUESTION 11

17 {c}^{3}  {d}^{2}  = 17 \times  {c}^{3}  \times  {d}^{2}

5 {c}^{2}  {d}^{2}  = 5 \times  {c}^{2}  \times  {d}^{2}

GCF= {c}^{2}  \times  {d}^{2}

GCF= {c}^{2}{d}^{2}

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What is the equation of a line passing through (5,2) and parallel to the line represented by the equation y - 2x + 1?
34kurt

Step-by-step explanation:

Im not sure if its supposed y = -2x + 1 or y = 2x + 1 but I'll solve the problem for both.

First for y = -2x + 1

Since the line has to be parallel to y = -2x + 1

the slopes would be the same.

so, so far the equation would be

y = -2x + b

now we substitute (5,2) into the equation

2 = -2(5) + b

2 = -10 + b (Add 10 to both sides of the equation)

+10 +10

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The equation would be

y = -2x + 12

^^ This equation is parallel to y = -2x + 1

Now to solve for an equation parallel to y = 2x + 1

Both equations would have the same slope

So far we would have

y = 2x + b

Now we solve for be by substituting the point (5,2)

2 = 2(5) + b

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julia-pushkina [17]

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We will use the above-mentioned property to get rid of the common shapes that exist on both sides. Here, the balance represents equality between both sides.

Consider that the pans are initially balanced. And both sides contain a red cube. So we can subtract or eliminate the red cube and the equality will still hold true.

Similarly, we can eliminate the blue sphere.

At this point, we have the purple pyramid and a yellow diamond on the left side, and on the right side, we have 4 pieces of yellow diamond.

Observe that one piece of yellow diamond is common to both sides, so it can also be eliminated from both sides.

After this, the left side will contain only the purple pyramid while the right side contains 3 yellow diamonds. Since we are always using the subtraction property to eliminate the shapes, the balance must be maintained throughout the process.

At last, there must exist a balance between the left side containing the purple pyramid only, and the right side containing 3 yellow diamonds.

Therefore, it can be concluded that one purple pyramid is equal to 3 yellow diamonds.

6 0
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