La la la la ah boom aye boom boom boom ha yalllllllllllllllllllll

Hannah places her home on a grid so that each unit square on the grid equates to 11 mile. She uses the fact that the major stree

Aneli [31]

Answer:

64.1

Step-by-step explanation:

You use Pythagorean theorem which is x^2 +y^2=z^2 (^ this means to the power of) so you would get 55^2+33^2=z^2. solving this would get you z=64.1404708433 or 64.1

8 0

2 years ago

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

Which could be used to evaluate the expression Negative 6 (4 and two-thirds)?

Firdavs [7]

Answer:

The answer is (Negative 6) (4) + (negative 6) (two-thirds)

Step-by-step explanation:

So basically, Negative 6x4= negative 24. Then you have to add Negative 6 x two-thirds. So 6 times two-thirds equal 4. So now you just add 24 and 4. And that equals 28. :)

6 0

2 years ago

Read 2 more answers

What are the vertical asymptote(s) of y= (x-6)/(x+8) (x-7)

White raven [17]

Answer:

x = -8 and x= 7

Step-by-step explanation:

recall that for a rational expression, the vertical asymptotes occur at x-values that causes the expression to become undefined. These occur when the denominator becomes zero.

Hence the asymptototes will occur in x-locations where the denominator , i.e

(x+8)(x-7) = 0

solving this, we get

(x+8) = 0 ----> x = -8

or

(x-7) = 0 ------> x = 7

hence the asymptotes occur x = -8 and x= 7

5 0

2 years ago

Read 2 more answers

Is 4 to 7 the same as 7 to 4 explain why it why not

tino4ka555 [31]

It’s not the same because if you simplify them there going to end up looking different (the answers) And they get a whole different do you denominator a numerator

6 0

3 years ago