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zhannawk [14.2K]
2 years ago
15

Five books and four pens cost $800. One book costs $120. How much does one pen cost?

Mathematics
1 answer:
cestrela7 [59]2 years ago
3 0
The cost of each pen is $50, because you spend $600 on books leaving $200 to buy FOUR pens, making it $50 per pen
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I don’t get it! It says “Match each division expression to it’s quotient.” So can anyone help with 1/12 / 1/6 ? Please!
natta225 [31]

We have two fractions being divided. When we divide two fractions, we flip the second fraction and multiply

1/12 divided by 1/6 = 1/12 * 6/1

-------

From here we multiply straight across.

The numerators multiply to get 1*6 = 6

The denominators multiply to 12*1 = 12

We end up with 6/12 which reduces to 1/2 when you divide both parts by the GCF 6

<h3>Final Answer:  1/2</h3>
5 0
3 years ago
How do you solve this equation? <br>2x+8-22= -2
Schach [20]

Answer:

x=6

Step-by-step explanation:

First you have to subtract 8 from 22 and you get -14. Then you rewrite the equation, 2x-14=-2.

Since your subtracting 14 you would add 14 to both sides, so -14+14 and -2+14. -2+14 will turn out to be 12 and then you divide 2 from both sides, so 2x-2 and 12-2.

Your answer will be x=6

4 0
2 years ago
Read 2 more answers
374 × 510<br><br> What are the partial products Lucas will need to solve the problem?
BabaBlast [244]
The partial products that Lucas would need to solve this problem would be 3,740 and 187,000 if you are multiplying 374 on the top column and 510 on the bottom column. Or, the partial products Lucas would need is 2,040 and 35,700 and 153,000. Both partial products combination added together would be 190,740.
6 0
3 years ago
X/3 - 38 =42 what is the value of x
madam [21]
X/3 - 38 = 42

x/3 = 80

x = 240
8 0
3 years ago
ALGEBRA QUESTION PLS HELPS
cluponka [151]

The value of x is –7.

Solution:

Given expression:

$\left(\frac{1}{x+3}+\frac{6}{x^{2}+4 x+3}\right) \cdot \frac{x+3}{x+1}

Let us factor x^2+4x+3.

x^2+4x+3=(x+1)(x+3)

Substitute this in the fraction.

$\left(\frac{1}{x+3}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

To make the denominator same, multiply and divide the first term by (x +1).

$\left(\frac{(x+1)}{(x+1)(x+3)}+\frac{6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

Denominators are same, you can add the fractions.

$\left(\frac{x+1+6}{(x+1)(x+3)}\right) \cdot \frac{x+3}{x+1}

$\frac{x+7}{(x+1)(x+3)} \cdot \frac{x+3}{x+1}

Cancel the common term in the numerator and denominator.

$\frac{x+7}{x+1} \cdot \frac{1}{x+1}

Multiply the fractions.

$\frac{x+7}{(x+1)^2}

$\frac{x+7}{x^2+2x+1}

The expression is simplified to one rational expression.

Suppose the expression is equal to 0.

$\frac{x+7}{x^2+2x+1}=0

Do cross multiplication.

${x+7}=0\times (}{x^2+2x+1})

Any number or variable multiplied by 0 gives 0.

${x+7}=0

Subtract 7 from both sides of the equation.

${x+7-7}=0-7

x = –7

The value of x is –7.

7 0
3 years ago
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