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3 years ago

11

How do u multiply monimials

1 answer:

krek1111 [17]3 years ago

6 0

Two multiply 2 monomials, you have to multiply the coefficients and add the degrees. For example

2x^2 × 3x^3 =

6x^5

You might be interested in

0.6 + 15b + 4 = 25.6 all equivelant

Mila [183]

Answer:

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

Step-by-step explanation:

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

3 0

3 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

What is the circumference, in centimeters, of the circle? Use 3.14 for π. Enter your answer in the box. (the diameter is 24cm)

gizmo_the_mogwai [7]

The circumference of the circle is 75.36 cm.

Step 1: Given data:

Diameter of the circle is 24 cm,

$\pi = 3.14$

Step 2: Formula used and calculation:

Circumference of the circle is $2\pi r$ .

So, Radius, r = 24/2 = 12 cm.

Now, Putting the value in the above formula, we get

2 × 3.14 ×12 = 3.14 × 24 = 75.36 cm.

Hence, the circumference of the circle is 75.36 cm.

Learn more about circumference of the circle, refer:

brainly.com/question/23964435

8 0

2 years ago

What is 0.1m + 8 - 12n when m = 30 and n= 1/4

tia_tia [17]

Answer:

8

Step-by-step explanation:

Replace your variables with the numbers given.

0.1m + 8 - 12n =

0.1 · 30 + 8 - 12 · 1/4 =

8

3 0

3 years ago

1. (a) Use the integral test to show that P[infinity] n=1 1/n4 converges. (b) Find the 10th partial sum, s10, of the series P[in

scZoUnD [109]

Answer:

Step-by-step explanation:

a) $\int\limits^{\infty} _1 {\frac{1}{n^4} } \, dn\\ =\frac{n^{-3} }{-3}$

Substitute limits to get

= $\frac{1}{3}$

Thus converges.

b) 10th partial sum =

$\int\limits^{10} _1 {\frac{1}{n^4} } \, dn\\ =\frac{n^{-3} }{-3}$

= $\frac{-1}{3} (0.001-1)\\= 0.333$

c) Z [infinity] n+1 1 /x ^4 dx ≤ s − sn ≤ Z [infinity] n 1 /x^ 4 dx, (1)

where s is the sum of P[infinity] n=1 1/n4 and sn is the nth partial sum of P[infinity] n=1 1/n4 .

(question is not clear)

3 0

3 years ago