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sattari [20]
3 years ago
12

Choose the correct simplification of the expression (2x^2y^6z^5)(5x^4y^5z^3)

Mathematics
2 answers:
Olegator [25]3 years ago
8 0

Answer:

10x^6y^{11}z^8

Step-by-step explanation:

The expression is:

(2x^2y^6z^5)(5x^4y^5z^3)

to simplify we need to multiply the coefficients (2*5) and also we need to multiply the variables with following rule:

a^m*a^n=a^{m+n}

since we have x^2 multiplied with x^4 we will have x^6, and so on with y and z.

Thus, we will get:

(2x^2y^6z^5)(5x^4y^5z^3) =2*5x^{2+4}y^{6+5}z^{5+3}

simplifying the expression and the exponents:

(2x^2y^6z^5)(5x^4y^5z^3) =10x^6y^{11}z^8

IgorLugansk [536]3 years ago
6 0
<span>2 x^2 y^6 z^5
5 x^4 y^5 z^3
---------------
10 ^6 ^11 ^8</span>
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Answer:

Factored form (linear): (4x+3)(x-3)(x-5)

Step-by-step explanation:

Using long division and factoring, we can find the answer. Since the root/zero of <em>5 </em>is given already, we can change this into a factor of <em>(x - 5)</em><em> </em>as 5 is a root.

Step 1: Long division. Divide f(x) by (x-5)

Once you do long division, you should have left over (x-5)(4x²+9x-9)

Step 2: Factor the quadratic

Since you can't do gcf, you have to leave the 4x as a factor. Once you factor out the quadratic, you should get the answer:

(x-5)(4x+3)(x-3)

Step 3: Rearrange (Optional)

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3 0
3 years ago
At the zoo tje ratio of snakes to lizards is 3:2 A. If there were 10 lizards ,how many snakes would be there be? B.If there were
yawa3891 [41]

Answer:

Part A) 15\ snakes

Part B) 6\ lizards

Part C) The zoo would need to get four more lizards to maintain the same proportion

Part D) 12 snakes and 8 lizards

Step-by-step explanation:

Part A) If there were 10 lizards ,how many snakes would be there be?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2} ----> equation A

we have that

y=10\ lizards

substitute the value of y in the equation A

\frac{x}{10}=\frac{3}{2}

solve for x

x=10(3)/2\\x=15\ snakes

Part B) If there were 9 snakes ,how many lizards would there be?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2} ----> equation A

we have that

x=9\ snakes

substitute the value of x in the equation A

\frac{9}{y}=\frac{3}{2}

solve for y

y=9(2)/3\\y=6\ lizards

Part C) If the number of snakes in the zoo is increased by 6, how many more lizards would the zoo need to get to keep the same ratio?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2} ----> equation A

For x=6\ snakes

substitute the value of x in the equation A

\frac{6}{y}=\frac{3}{2}

solve for y

y=6(2)/3\\y=4\ lizards

therefore

The zoo would need to get four more lizards to maintain the same proportion

Part D) If the total number of snakes and lizards at the zoo was 20, how many snakes and lizards would there be?

Let

x ---> the number of snakes

y ---> the number of lizards

\frac{x}{y}=\frac{3}{2}

isolate the variable x

x=1.5y ----> equation A

x+y=20 ----> equation B

solve the system by substitution

substitute equation A in equation B

1.5y+y=20

solve for y

2.5y=20

y=8\ lizards

Find the value of x

x=1.5(8)=12\ snakes

therefore

12 snakes and 8 lizards

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