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3 years ago

What is the solution to the inequality 13 x + 3 < 42 ?

Mathematics

1 answer:

zepelin [54]3 years ago

6 0

Answer:

x < -15

Step-by-step explanation:

Here is the solution:

-3x-42 > 3

Add the numbers

-3x > 45

Divide both sides of the inequation by -3 and flip the inequality sign.

x < -15

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3 years ago

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Simplify this please

Ugo [173]

Answer:

$\frac{12q^{\frac{7}{3}}}{p^{3}}$

Step-by-step explanation:

Here are some rules you need to simplify this expression:

Distribute exponents: When you raise an exponent to another exponent, you multiply the exponents together. This includes exponents that are fractions. $(a^{x})^{n} = a^{xn}$

Negative exponent rule: When an exponent is negative, you can make it positive by making the base a fraction. When the number is apart of a bigger fraction, you can move it to the other side (top/bottom). $a^{-x} = \frac{1}{a^{x}}$ , and to help with this question: $\frac{a^{-x}b}{1} = \frac{b}{a^{x}}$ .

Multiplying exponents with same base: When exponential numbers have the same base, you can combine them by adding their exponents together. $(a^{x})(a^{y}) = a^{x+y}$

Dividing exponents with same base: When exponential numbers have the same base, you can combine them by subtracting the exponents. $\frac{a^{x}}{a^{y}} = a^{x-y}$

Fractional exponents as a radical: When a number has an exponent that is a fraction, the numerator can remain the exponent, and the denominator becomes the index (example, index here ∛ is 3). $a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = (\sqrt[n]{a})^{m}$

$\frac{(8p^{-6} q^{3})^{2/3}}{(27p^{3}q)^{-1/3}}$ Distribute exponent

$=\frac{8^{(2/3)}p^{(-6*2/3)}q^{(3*2/3)}}{27^{(-1/3)}p^{(3*-1/3)}q^{(-1/3)}}$ Simplify each exponent by multiplying

$=\frac{8^{(2/3)}p^{(-4)}q^{(2)}}{27^{(-1/3)}p^{(-1)}q^{(-1/3)}}$ Negative exponent rule

$=\frac{8^{(2/3)}q^{(2)}27^{(1/3)}p^{(1)}q^{(1/3)}}{p^{(4)}}$ Combine the like terms in the numerator with the base "q"

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)}q^{(1/3)}}{p^{(4)}}$ Rearranged for you to see the like terms

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)+(1/3)}}{p^{(4)}}$ Multiplying exponents with same base

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(7/3)}}{p^{(4)}}$ 2 + 1/3 = 7/3

$=\frac{\sqrt[3]{8^{2}}\sqrt[3]{27}p\sqrt[3]{q^{7}}}{p^{4}}$ Fractional exponents as radical form

$=\frac{(\sqrt[3]{64})(3)(p)(q^{\frac{7}{3}})}{p^{4}}$ Simplified cubes. Wrote brackets to lessen confusion. Notice the radical of a variable can't be simplified.

$=\frac{(4)(3)(p)(q^{\frac{7}{3}})}{p^{4}}$ Multiply 4 and 3

$=\frac{12pq^{\frac{7}{3}}}{p^{4}}$ Dividing exponents with same base

$=12p^{(1-4)}q^{\frac{7}{3}}$ Subtract the exponent of 'p'

$=12p^{(-3)}q^{\frac{7}{3}}$ Negative exponent rule

$=\frac{12q^{\frac{7}{3}}}{p^{3}}$ Final answer

Here is a version in pen if the steps are hard to see.

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So first we need to find what percent of the students walk to school.

50 + 35 = 85% don't walk leaving only 15% that walk to school,

To find the 15% percent that DO walk we multiply .15 by the total number of people.

.15 x 1640 = 246 people walk to school

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Ted's take-home pay is $1900 a month. He spends 17% of his take-home pay on groceries. How much do groceries cost Ted each month

Licemer1 [7]

$323 . just multiply 1900 by 0.17

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