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1 year ago

Find the slope of the line that passes through (5,12) and (9,7)

Mathematics

1 answer:

omeli [17]1 year ago

8 0

The slope will be the difference between the y-coordinates divided by the difference between the x-coordinates:

then the slope of the line is:

$m=\frac{12-7}{5-9}=\frac{5}{-4}=-\frac{5}{4}$

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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.

5 0

2 years ago

f(x) = 3 cos(x) 0 ≤ x ≤ 3π/4 evaluate the Riemann sum with n = 6, taking the sample points to be left endpoints. (Round your ans

Kruka [31]

Answer:

$\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558$

Step-by-step explanation:

We want to find the Riemann sum for $\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx$ with n = 6, using left endpoints.

The Left Riemann Sum uses the left endpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)$

where $\Delta{x}=\frac{b-a}{n}$ .

Step 1: Find $\Delta{x}$

We have that $a=0, b=\frac{3\pi }{4}, n=6$

Therefore, $\Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}$

Step 2: Divide the interval $\left[0,\frac{3 \pi}{4}\right]$ into n = 6 sub-intervals of length $\Delta{x}=\frac{\pi}{8}$

$a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b$