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3 years ago

Please need help with this math question

Mathematics

2 answers:

Tomtit [17]3 years ago

5 0

Answer:

third option

Step-by-step explanation:

We just have to calculate 2x² - 4x - (x² + 6x). 2x² - x² = x² and -4x - 6x = -10x so the answer is x² - 10x.

True [87]3 years ago

3 0

Answer:

x^2-10x

Step-by-step explanation:

f(x)-g(x)

(2x^2-4x)-(x^2+6x)

carry through the negative

2x^2-4x-x^2-6x

x^2-10x

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Is my answer right to this problem

VLD [36.1K]

Yes it’s right for this problem

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2 years ago

Given 3 sides of a triangle - how to tell it right, acute, or obtuse 8,11, 16

user100 [1]

Using Pythagorean theorem, we can justify our answer.
if c^2 = a^2 + b^2 then triangle is right one,
if c^2 > a^2 + b^2 then triangle is obtuse and
if c^2 < a^2 + b^2 then triangle is acute triangle.

Here a=8, b=11 and c=16
Put these values in the equation
16^2 = 8 ^2 + 11^2
256 = 64 +121
256> 185
So here c^2 > a^2 + b^2 Which means triangle is obtuse triangle.

Answer: Obtuse Triangle

6 0

3 years ago

boeing plans to increase its price for jetliners. with a selling price $201.5 million and a cost of $190.1 million, what was the

scoray [572]

so from 190.1 to 201.5 is 201.5 - 190.1 = 11.4.

now, if we take 190.1 as the 100%, what is 11.4 off of it in percentage?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 190.1&100\\ 11.4&x \end{array}\implies \cfrac{190.1}{11.4}=\cfrac{100}{x}\implies 190.1x=1140 \\\\\\ x=\cfrac{1140}{190.1}\implies x\approx 5.99684$

3 0

3 years ago

Read 2 more answers

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.