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2 years ago

Three more than twice a number is less than negative thirteen. Solve the inequality for the unknown number.

Mathematics

1 answer:

Goshia [24]2 years ago

4 0

Answer:

Step-by-step explanation:

The number = x

Three more than the number = x + 3

twice the number = 2x

which means:

2x + 3 < -13

2x + 3 -3 < -13 -3 (subtract 3 from both sides)

2x < -16

2x/2 < -16/2 (divide 2 from both sides)

x < -8

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Juan makes 75% of the goals he attempted during soccer practice. If he makes 30 goals at practice, how many does he attempt?

m_a_m_a [10]

Answer:

22.5

Step-by-step explanation:

30/4=7.5

7.5 times 3 = 22.5

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5 0

2 years ago

Simplify the remaining expression completely x^6y^4/xy^2

polet [3.4K]

Answer: x^5y^2

Step-by-step explanation:

6 0

3 years ago

The answer and how you got the answer

olasank [31]

Answer:

$\frac{34}{15}$

Step-by-step explanation:

we are asked to evaluate

$4\tfrac{1}{4}$ ÷ $1\tfrac{7}{8}$

Above we are given mixed fraction which can be converted in proper fraction using formula given below

$a\tfrac{b}{c}=\frac{a \times c +b}{c}$

Hence

$4\tfrac{1}{4}=\frac{4 \times 4 +1}{4}$

$4\tfrac{1}{4}=\frac{17}{4}$

Also

$1\tfrac{7}{8}=\frac{1 \times 8 +7}{8}$

$1\tfrac{7}{8}=\frac{15}{8}$

Hence

$4\tfrac{1}{4}$ ÷ $1\tfrac{7}{8}$

can be written as

$\frac{17}{4}$ ÷ $\frac{15}{8}$

Also we know the rule for dividing fraction is as given below

$\frac{a}{b}$ ÷ $\frac{c}{d}$ = $\frac{a}{b} \times \frac{d}{c}$

Hence

$\frac{17}{4}$ ÷ $\frac{15}{8}$ = $\frac{17}{4} \times \frac{8}{15}$

= $\frac{17 \times 2}{15}$

= $\frac{34}{15}$

8 0

3 years ago

Can you please help me answer this question what's a+3

Inessa [10]

Answer:

a+3

Step-by-step explanation:

You cannot go any further in answering this question. These two terms are not like terms so they cannot be combined. Therefore, the answer is just a+3 itself.

7 0

3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago