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3 years ago

The cube of the sum of 4 and 9 times x divided by the product of 5 times x and the difference of x and 1

1 answer:

5 0

+. Add x to y x + y y added to 7. 7+ y. The sum of a and b a + b m more than n ... 9 plus m. 9 + m. Subtraction. Subtract, subtract from, difference, between, less, less ... 7 times y. 7y. The product of x and y xy. 5 multiplied by y. 5y one-fifth of p. 1. 5 p ... Power. The square of y y. 2. The cube of k k. 3 t raised to the fourth power t. 4.

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

Use decimals and fractions in the same equation showing the Commutative Property. Repeat for the Associative Property.

Anna71 [15]

For Commutative Property of Addition

Let us take a decimal number $2.14$ and a fraction $\frac{5}{20}$

Now, according to the Commutative property of Addition:

For any two numbers $a$ and $b$ :

$a+b = b+a$

So, for $2.14$ and $\frac{5}{20}$

Let us add

$2.14+\frac{5}{20} = \frac{214}{100} +\frac{5}{20}$

$=\frac{214}{100} + \frac{5 \times 5}{20 \times 5} \\ \\=\frac{214}{100} + \frac{25}{100} \\ \\= 2.14 + 0.25 \\ \\ = 2.39$

Also,

$\frac{5}{20} + 2.14 = \frac{5}{20} + \frac{214}{100}$

$= \frac{5 \times 5}{20 \times 5} +\frac{214}{100} \\ \\= \frac{25}{100} + \frac{214}{100} \\ \\= 0.25 + 2.14 \\ \\ = 2.39$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

Hence, Commutative Property of Addition is satisfied.

For Associated Property of Addition

Let us take two same decimal numbers $2.14$ , $7.25$ and a fraction $\frac{5}{20}$

Now, according to the Associated property of Addition:

For any three numbers $a$ , $b$ and $c$

$a + (b+c) =(a+b) + c$

So, for $2.14$ , $7.25$ and $\frac{5}{20}$

The Left hand side:

$a + (b+c)$

$2.14 + (7.25 + \frac{5}{20}) = 2.14 + (\frac{725}{100} + \frac{5 \times 5}{20\times 5})$

$= 2.14 + (\frac{725}{100} + \frac{25 }{100})$

$= 2.14 + (\frac{750}{100} )$

$= \frac{214}{100} + \frac{750}{100}$

$= \frac{964}{100}$

$=9.64$

The Right hand side:

$(a + b )+c$

$(2.14 + 7.25 )+ \frac{5}{20}= ( 9.39 ) + \frac{5 }{20}$

$= 9.39 + \frac{5 \times 5 }{20 \times 5}$

$= 9.39 + \frac{25}{100}$

$= 9.39 + 0.25$

$= 9.64$

Thus,

$(2.14 + (7.25 + \frac{5}{20} )= (2.14 + 7.25 )+ \frac{5}{20}$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

Hence, Associative Property of Addition is satisfied.

8 0

3 years ago

In isosceles triangle ABC, AC=BC=20, the measure of angle A=68, and CD is the altitude to side AB. What is CD to the nearest 10t

antiseptic1488 [7]

Answer:

Step-by-step explanation:

Comment

The way the diagram looks when it is drawn is that you are going to need the sine function to solve for the side opposite which is CD.

Formula

Sin(A) = Sin(68 = opposite / hypotenuse Substitute values.

Solution

sin(68) = altitude (side opposite) / 20 multiply both sides by 20

20*sin(68) = 20* altitude / 20 substitute for sin(68)

20* 0.9272 = altitude

altitude = 18.5

Answer: 18.5 to the nearest tenth

3 0

2 years ago

Please help asap!!

masya89 [10]

If I were solving this equation, I would put x^2-20x+(100)=(10). Since this is in standard form, you could change this into vertex form by making the equation y=(x-10)^2 -10.

5 0

3 years ago

M(5,7) is the midpoint of RS the coordinates of S are (6,9) what are the coordinates of R

MaRussiya [10]

Answer:

(4,5) i believe

7 0

3 years ago

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