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3 years ago

The degree for 6g²h³k?

Mathematics

1 answer:

NeX [460]3 years ago

3 0

Answer:

the degree for this monomial is 7

Step-by-step explanation:

g^2 = 2

h^3= 3

k = 1

add the exponents- 2+3+1= 7

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<img src="https://tex.z-dn.net/?f=2x%20-%206%20%3D%205y%20-%203x" id="TexFormula1" title="2x - 6 = 5y - 3x" alt="2x - 6 = 5y - 3

Zigmanuir [339]

Answer:

( 2x - 3 ) / 5

Step-by-step explanation:

2x - 6 = 5y - 3

2x - 6 + 3 = 5y

2x - 3 = 5y

y = ( 2x - 3 ) / 5

6 0

2 years ago

PLEASE HELP!!!

AlekseyPX

Answer:

Step-by-step explanation:

Find the average rate of change of each given function over the interval [-2, 2]]:

✔️ Average rate of change of m(x) over [-2, 2]:

Average rate of change = $\frac{m(b) - m(a)}{b - a}$

Where,

a = -2, m(a) = -12

b = 2, m(b) = 4

Plug in the values into the equation

Average rate of change = $\frac{4 - (-12)}{2 - (-2)}$

= $\frac{16}{4}$

Average rate of change = 4

✔️ Average rate of change of n(x) over [-2, 2]:

Average rate of change = $\frac{n(b) - n(a)}{b - a}$

Where,

a = -2, n(a) = -6

b = 2, n(b) = 6

Plug in the values into the equation

Average rate of change = $\frac{6 - (-6)}{2 - (-2)}$

= $\frac{12}{4}$

Average rate of change = 3

✔️ Average rate of change of q(x) over [-2, 2]:

Average rate of change = $\frac{q(b) - q(a)}{b - a}$

Where,

a = -2, q(a) = -4

b = 2, q(b) = -12

Plug in the values into the equation

Average rate of change = $\frac{-12 - (-4)}{2 - (-2)}$

= $\frac{-8}{4}$

Average rate of change = -2

✔️ Average rate of change of p(x) over [-2, 2]:

Average rate of change = $\frac{p(b) - p(a)}{b - a}$

Where,

a = -2, p(a) = 12

b = 2, p(b) = -4

Plug in the values into the equation

Average rate of change = $\frac{-4 - 12}{2 - (-2)}$

= $\frac{-16}{4}$

Average rate of change = -4

The answer is D. Only p(x) has an average rate of change of -4 over [-2, 2]

3 0

3 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago

What is the value of the rational expression below when xis equal to 4?

horrorfan [7]

The answer is the letter c. 3

4 0

3 years ago

Find the area of the following circles. a. A circle with a 8-inch radius b. A circle with a 10-kilometer radius c. A circle with

Thepotemich [5.8K]

Area of a circle is $A= \pi r^2$ . The first one, a, the radius squared is 8*8=64. So $A=64 \pi$ . If you multiply in pi as 3.14, you'll have 200.96 square inches. The second one, b, the radius squared is 10*10=100. So $A=100 \pi$ square kilometers. If you multiply in pi as 3.14 you'll have 314 square kilometers. The third one, c, the radius squared is 14*14=196. So $A=196 \pi yd^2$ . Or you could multiply in pi as 3.14 to get 615.44 yd squared. For the last one, d, the radius squared is 22*22=484 cm. Therefore, $A=484 \pi cm^2$ , or multiply in 3.14 for pi to get 1519.76 cm squared. There you go!

5 0

3 years ago

Read 2 more answers