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diamong [38]
2 years ago
8

Factor 5w + 20 please and show your work. I need help

Mathematics
2 answers:
mars1129 [50]2 years ago
7 0

The factored version of the above statement would be 5(x + 4)

In order to find this, you need to find the greatest common factor of the two coefficients. First, list the factors of each.

Factors of 5: 1, 5

Factors of 20: 1, 2, 4, 5, 10, 20

Since the highest that exists in both lists is 5, we can divide both terms by 5 and pull it out of the parenthesis like this:

5(w + 4)

Which is your final answer

lesya692 [45]2 years ago
4 0

Answer: 25w

Step-by-step explanation:

5w+20

5+20=25

25w

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a coin operated machine sells plastic rings . it contains 6 yellow rings,11 blue rings,15 green rings,and 3 black rings.sarah pu
Mariana [72]
Sample space ={6y+11b+15g+3bk} = 35 elements
number of black rings=3
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5 0
3 years ago
A) In a group of 60 students, 15 liked maths only, 20 liked science only and 5 did not like
Paladinen [302]

Part (i)

We have 60 students total, and 5 didn't like any of the two subjects, so that must mean 60-5 = 55 students liked at least one subject.

<h3>Answer: 55</h3>

=========================================================

Part (ii)

We have 15 who like math only, 20 who like science only, and 55 who like either (or both). Let x be the number of people who like both classes.

We can then say

15+20+x = 55

x+35 = 55

x = 55-35

x = 20

This means 20 people liked both subjects

<h3>Answer: 20</h3>

=========================================================

Part (iii)

There are 15 people who like math only, and 20 who like both. Therefore, there are 15+20 = 35 people who like math (and some of these people also like science)

<h3>Answer: 35</h3>

=========================================================

Part (iv)

We'll follow the same idea as the previous part. There are 20 people who like science only and 20 who like both subjects. That yields 40 people total who like science (and some of these people also like math).

<h3>Answer: 40</h3>

=========================================================

Part (v)

We'll draw a rectangle to represent the entire group of 60 students. This is considered the universal set. Inside the rectangle will be two overlapping circles to represent math (M) and science (S).

We'll have 15 go in circle M, but outside circle S to represent the 15 people who like math only. Then we have 20 go in circle S but outside circle M to show the 20 people who like science only. We have another copy of 20 go in the overlapped region between the circles. This is the 20 people who like both classes. And finally, we have 5 go outside both circles, but inside the rectangle. These are the 5 people who don't like either subject.

Note how all of the values in the diagram add up to 60

15+20+20+5 = 60

This helps confirm we have the correct values.

<h3>Answer: See the venn diagram below</h3>

3 0
3 years ago
the sum of two consecutive integers is -49 write an equation that models this situation and find the values of the two integers
hodyreva [135]
Х is the first <span>integer
</span>(x+1) is the second integer

x + x + 1 = -49
2x + 1 = - 49
2x = -49 - 1
2x = -50
x = -25  first integer
-25 + 1 = -24 second integer

Answer: -25 and -24
7 0
3 years ago
Read 2 more answers
Perform the indicated operation. Be sure the answer is reduced.
avanturin [10]
<h3>Given Equation:-</h3>

\boxed{ \rm  \frac{4x^{2}y^{3}z}{9} \times  \frac{45y}{8 {x}^{5} {z}^{5} }}

<h3>Step by step expansion:</h3>

\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times  \dfrac{45y}{8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times  \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times  \dfrac{45y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times  \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}z}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5y \times  {y}^{3} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5y {}^{0}  \times  {y}^{3 + 1} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5 \times  {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \bf  \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}

8 0
2 years ago
I need help solving this, it has 2 parts.
Ratling [72]

Answer:

the part of second is 4second p

6 0
3 years ago
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