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Galina-37 [17]
3 years ago
13

Solve the equation 3x + 2 = 4x + 5 using algebra tiles. Which tiles need to be added to both sides to remove the smaller coeffic

ient? 3 positive x-tiles3 negative x-tiles 4 positive x-tiles 4 negative x-tiles Which tiles need to be added to both sides to remove the constant from the right side of the equation? 2 positive unit tiles 2 negative unit tiles 5 positive unit tiles 5 negative unit tiles What is the solution? x = –3x = –1x = 3x = 7
Mathematics
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

Step-by-step explanation:

Given the expression

3x + 2 = 4x + 5

1. The smaller coefficient of x is 3

to remove the smaller coefficient

we need to add - 3x to both sides

3x +2 + (-3x) = 4x +5 (-3x)

3x + 2 - 3x = 4x + 5 - 3x

Collecting like terms we have

3x-3x+2= 4x-3x+5

2 = x+ 5

2. The constant on the right side is

5,to remove the constant from the right side of the equation we need to add - 5 to both sides

3x + 2+ (- 5) = 4x + 5 +(-5)

3x+ 2-5 =4x +5-5

3x-3 = 4x

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What is the answer?please help me
Maslowich

Answer:

y = 2x - 1 Graph B

y = 2x + 2 Graph A

y = 2x - 5 Graph C

Step-by-step explanation:

y = mx + c

c is where the line intersect the Y axis.

<h2>Substitute x for 0:</h2>

y = 2x - 1

2 * 0 - 1 = -1.

This is graph B because it intersects at 0,-1.

y = 2x + 2

2 * 0 + 2 = 2

This is graph A because it intersects at 0,2.

y = 2x - 5

2 * 0 - 5 = -5

This is graph C because it intersects at 0,-5.

6 0
3 years ago
Jason inherited a piece of land from his great-uncle. Owners in the area claim that there is a 45% chance that the land has oil.
Anna [14]
In this case you the probabilities can be multiplied similarly to if you flipped a coin twice. .45 * .80 = .36
5 0
3 years ago
Read 2 more answers
Suppose both factors in a multiplication problem are multiples of 10. Why might the number of zeroes in the product be different
Lelu [443]

Answer:

Multiplying factors whose products are multiples of 10 add to the number of zeros when factors are multiplied as multiples of 10s.

Step-by-step explanation:

Let the first five multiples of ten be

10*1= 10

10*2= 20

10*3=30

10*4=40

10*5= 50

Suppose we chose 20 and 50.

Now multiplying 20 with 50 we get

20*50= 1000

IF we count the total number of zeros in the factors ( 20 and 50) they are 2.

But the number of zeros in the product (1000) are 3.

This is because when we multiply 2 with 5 we get 10 which adds to the existing number of zeros ( i.e 2) and we get a total of 3 zeros.

And multiplying 10 with 50 we get

10*50= 500

IF we count the total number of zeros in the factors ( 10 and 50) they are 2.

But the number of zeros in the product (500) are also 2.

This is because when we multiply 1 with 5 we get 5 which does not add to the existing number of zeros ( i.e 2) and the total number of zeros remain the same.

Similarly multiplying 20 with 30 we get

20*30= 600

IF we count the total number of zeros in the factors ( 20 and 30) they are 2.

But the number of zeros in the product (600) are also 2.

This is because when we multiply 2 with 3 we get 6 which does not have a zero and the total  number of zeros remain the same as in the factors.

So we see that multiplying factors whose products are multiples of 10 add to the number of zeros when factors are multiplied as multiples of 10s.

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B80%7D%20" id="TexFormula1" title=" \sqrt[3]{80} " alt=" \sqrt[3]{80} " ali
Marianna [84]

Start by decomposing the number inside the root into primes

Then group the terms into cubes if possible

\begin{gathered} 80=2\cdot2\cdot2\cdot2\cdot5 \\ 80=2^3\cdot2\cdot5 \\ 80=10\cdot2^3 \end{gathered}

rewrite the root

\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}

then cancel the terms that are cubes and bring them out of the root

\sqrt[3]{80}=2\sqrt[3]{10}

7 0
1 year ago
How many times greater is the 4 in the thousands place compared to the 4 in the hundreds place?
lora16 [44]
It is 10 more in the thousands than the hundreds... multiply 100 x 10...
4 0
3 years ago
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