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3 years ago

Greatest common factor of −27x2yz5 + 15x3z3

1 answer:

8 0

The GCF will be found as follows:
-27x^2yz^5+15x^3z^3
-27x^2yz^5=-3*3*3*x*x*y*z*z*z*z*z
15x^3z^3=3*5*x*x*x*z*z*z
the GCF is the product of the lowest power of each factor that appears in each term
thus we shall have:
3*z*z*z*x*x
=3z^3x^2

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What is 1.86 divided by 0.15 and why?

ch4aika [34]

Answer:

The answer is 12.4

Step-by-step explanation:

Because it is this way.

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3 years ago

Solve the absolute value equation.<br><br> 4 − |x| = 1

Alika [10]

Answer:

x = +3 and -3

Step-by-step explanation:

Start by solving for |x| as normal:

4 - |x| = 1

- |x| = -3 (subtract 4 from both sides)

|x| = 3 (multiply both sides by -1)

the absolute value of x means it will be positive, so you need to recognize that whatever is in the absolute value bars will be made positive.

this means that x = 3 AND -3, because both |3| and |-3| = 3

check your answer by plugging x = 3 and x = -3 back in:

4 - |3| = 1

4 - 3 = 1

1 = 1

and

4 - |-3| = 1

4 - 3 = 1

1 = 1

5 0

2 years ago

Read 2 more answers

Al sunrise, the temperature outside was 5 below zero By noon the temperature rose 27°. What was the temperature at noon?

irina1246 [14]

Answer:

the temperature at noon was 22 can i get brainliest please? :)

Step-by-step explanation:

-5 + 27 = 22

7 0

4 years ago

2. What is 0.03249 in scientific notation?

Afina-wow [57]

Answer:

3.249 X 10^-2

Step-by-step explanation:

$\huge 0.03249 = 3.249 \times {10}^{ - 2} \\$

$\huge 715 \times {10}^{3} = 715000 \\$

6 0

4 years ago

32. Parallelogram ABCD has vertices A(0,0), B(2,4), and C(10,4). Find the coordinates of D.

goblinko [34]

Answer:

The fourth vertex is D(8, 0)

Step-by-step explanation:

Let A(0, 0), B(2, 4), and C(10, 4) be the three vertices of a parallelogram ABCD and let its fourth vertex be D(a, b).

Join AC and BD. Let AC and BD intersect at point O.

We have known that the diagonals of a parallelogram bisect each other.

So, O would be the midpoint of AC as well as that of BD.

$A\left(0,\:0\right)=\left(x_1,\:x_2\:\right)$

$C\left(10,\:4\right)=\left(x_2,\:x_2\:\right)$

The midpoint of AC is:

$\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$

$\left(x_1,\:y_1\right)=\left(0,\:0\right),\:\left(x_2,\:y_2\right)=\left(10,\:4\right)$

$=\left(\frac{10+0}{2},\:\frac{4+0}{2}\right)$

$=\left(5,\:2\right)$

The midpoint of BD is:

$=\left(\frac{a+2}{2},\:\frac{b+4}{2}\right)$

∵ $\frac{a+2}{2}=5$

$a+2=10$

$a=8$

∵ $\frac{b+4}{2}=2$

$b+4=4$

$b=0$

Hence, the fourth vertex is D(8, 0)

4 0

3 years ago