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3 years ago

CAN I GET HELP ON THIS PLEASE WILL GIVE 5 STARS

Mathematics

1 answer:

nikklg [1K]3 years ago

6 0

Answer:

I don't know the ans please search on the Google you will get

And don't forget to mark me as brainlest please guys and follow me back please please

You might be interested in

Find the missing coefficient (4d—4)(3d—2)=12d^2+ []d+8

sergij07 [2.7K]

<span>(4d-4)(3d-2)=12d^2 + []d + 8

Using FOIL method:
4d x 3d = 12d^2
4d x -2 = -8d
-4 x 3d = -12d
-4 x -2 = 8

Adding common terms, we have:
-8d + -12d = -20 d

The missing coefficient is -20.</span>

8 0

3 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

3 years ago

Please help ASAP!!!!!!!!!!

zepelin [54]

Answer:

(c-a, 0)

Step-by-step explanation:

The horizontal space between (c, b) and P is the same as the space between (a, b) and O.

Coordinates are written (x, y), where x is for horizontal space.

P is on the x-axis, making the y-coordinate 0.

(a+c, 0) would be to the right of the entire parallelogram.

(c, 0) would be directly below (c, b).

(a-c, 0) would be to the left of the entire parallelogram and in the other quadrant.

4 0

3 years ago

In a certain elementry school, 52% of the students are girls. a sample of 65 students is drawn. what is the probability that mor

Rainbow [258]

Answer:

12432435

Step-by-step explanation:

gvngxsfgjghdjmhg

7 0

3 years ago

Read 2 more answers

A set of weights includes a 5 lb barbell and 4 pairs of weight plates. Each pair of

Zina [86]

Answer:

The answer to this question would be B:

Based on the question, since the weight of the weight plates are 20 lbs, this would be represented by the 20x in the function. As well, the 5 lb barbell would be represented by the 5 in the function. The range of the function is determined by the amount of weight plates are added. So if I added one weight plate the equation would equal, f(x) = 20(1) + 5 = 25. This continues on the more and more weight plates you add.

Hope this reached you well :)

Step-by-step explanation:

20 = weight of the weight plates

x = amount of weight plates.

5 = weight of the barbell

f(x) = 20(0) + 5 = 5

f(x) = 20(1) + 5 = 25

f(x) = 20(2) + 5 = 45

f(x) = 20(3) + 5 = 65

f(x) = 20(4) + 5 = 85

5 0

3 years ago