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3 years ago

Factor the common factor out of each expression

Mathematics

1 answer:

Dvinal [7]3 years ago

4 0

Its 15*5 ................................................................................

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Match the expression on the left with the correct simplified expression on the right.

andriy [413]

Given: The expression below

$\begin{gathered} (\frac{(3x^3y^4)^3}{(3x^2y^2)^2})^2 \\ (\frac{(3x^4y^2)^4}{(3x^5y^2)^3})^2 \end{gathered}$

To Determine: The matching expression to the given expressions

Solution

Let us simplify each of the expressions using exponents rule

$\begin{gathered} Exponent-Rule1=(a^m)^n=a^{m\times n} \\ Exponent-Rule2=(\frac{a^m}{a^n})=a^{m-n} \end{gathered}$

Applying the exponent rule 1 above to the given expressions

$\begin{gathered} (3x^3y^4)^3=3^3x^{3\times3}y^{4\times3}=27x^9y^{12} \\ (3x^2y^2)^2=3^2x^{2\times2}y^{2\times2}=9x^4y^4 \end{gathered}$ $\begin{gathered} (3x^4y^2)^4=3^4x^{4\times4}y^{2\times4}=81x^{16}y^8 \\ (3x^5y^2)^3=3^3x^{5\times3}y^{2\times3}=27x^{15}y^6 \end{gathered}$

Applying the exponent rule 2

$\frac{(3x^{3}y^{4})^{3}}{(3x^{2}y^{2})^{2}}=\frac{27x^9y^{12}}{9x^4y^4}=\frac{27}{9}x^{9-4}y^{12-4}=3x^5y^8$ $\frac{(3x^{4}y^{2})^{4}}{(3x^{5}y^{2})^{3}}=\frac{81x^{16}y^8}{27x^{15}y^6}=\frac{81}{27}x^{16-15}y^{8-6}=3xy^2$

Let us not apply exponent rule 1 above

$(\frac{(3x^{3}y^{4})^{3}}{(3x^{2}y^{2})^{2}})^2=(3x^5y^8)^2=3^2x^{5\times2}y^{8\times2}=9x^{10}y^{16}$ $(\frac{(3x^{4}y^{2})^{4}}{(3x^{5}y^{2})^{3}})^2=(3xy^2)^2=3^2x^2y^{2\times2}=9x^2y^4$

Hence, the matching is as shown below

8 0

2 years ago

Sue can run 18 miles in 2 hours and can bike 60 miles in 3 hours what percent of sue average biking speed is her average running

cupoosta [38]

18/2 = 9 Miles per hour for running

60/3 = 20 Miles per hour for biking

7 0

4 years ago

Read 2 more answers

What algebraic expression must be added to the sum of

user100 [1]

Answer:

We need to add the expression: $4x^2-16$

Step-by-step explanation:

As the first step, add the first two polynomials listed in order to find what their sum is. Do such by combining like terms:

$3x^2+4x+8\,+ (2x^2-6x+3) =5x^2-2x+11$

now, you need to find another polynomial (let's call it P(x)) such that added to the one we just found above, will render $9x^2-2x-5$ as a result.

Putting this phrase in mathematical terms, we write:

$5x^2-2x+11\,+ P(x) = 9x^2-2x-5$

Now we solve for this unknown P(x) in this equation, by subtracting each of the terms that are accompanying it on the left side of the equation. We do one at a time for clarity:

$5x^2-2x+11\,+ P(x) = 9x^2-2x-5\\-2x+11+P(x)=9x^2-5x^2-2x-5\\-2x+11+P(x)=4x^2-2x-5\\11+P(x)=4x^2-2x+2x-5\\11+P(x)=4x^2-5\\P(x)=4x^2-5-11\\P(x)=4x^2-16$

Therefore, the expression we need to add is: $4x^2-16$

8 0

4 years ago

What is it a ,b,c,or D can I get a step by step explanation

ANEK [815]

The answers are b and c. Plug in -4 for x and solve each option

2(-4) -8 = 0. 0 does not equal 1
3(-4) +8 = -4
1/2 * -4 + 3 = 1, -4+5 = 1
2(1- -4) = 10, -4 + 2 = -2

4 0

4 years ago

In order to qualify for finals in racing event,a race car must achieve an average speed of 250km/hr on a track with a total leng

Kazeer [188]

Answer:

The minimum average speed needed in the second half is 270 km/hr

Step-by-step explanation

We can divide the track in two parts. For the first half of the track the average speed the car achieved was 230 km/hr and we need to make sure that the average speed of the full track is 250 km/hr. Then, we can calculate the average speed of the two parts of the track and force this to be equal to 250 km/hr. In equation, defining $v_2$ as the average speed of the second half:

$\frac{230 + v_2}{2}=250$

Solving for $x$

$x=250*2-230\\x=500-230\\x=270$

Therefore, achieving a speed of 270 km/hr in the second half would be enough to achieve an average speed of 250 on the track.

6 0

3 years ago