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3 years ago

Who wanna be friends?

Mathematics

2 answers:

Elan Coil [88]3 years ago

8 0

Answer:

yessssssss

Step-by-step explanation:

zhannawk [14.2K]3 years ago

5 0

Answer:

okkkkkkkkkkkkkkkkkkkkkk

Step-by-step explanation:

You might be interested in

Multiply 1/8 x 3/4 =

Novay_Z [31]

Answer:

3/32

Step-by-step explanation:

1 times 3= 3 and 8 times 4=32

8 0

3 years ago

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There are 10 guest attending the gathering. There are 5 turkeys and each turkey weighs 4 pounds. If each person eaf the same amo

Westkost [7]

I think you have multiply 5x4 which equals 20 then 20 divided by 10 which equals 2 and that’s ur answer

5 0

3 years ago

. Only 2% of a large population of 100-ohm gold-band resistors have resistances that exceed 105 ohms. a. For samples of size 100

Soloha48 [4]

Using the information given above, the sampling distribution of the sample proportion of 100-ohm gold-band is 2.

Sampling distribution of proportion, P = 2% = 0.02

Sample size, n = 100

The sampling distribution of the sample proportion can be calculated thus:

Distribution of sample proportion = np

Distribution of sample proportion = (100 × 0.02) = 2

Therefore, there is a probability that only 2 of the samples will have resistances exceeding 105 ohms.

Learn more : brainly.com/question/18405415

4 0

2 years ago

Find the sum of the first 48 terms of the following series, to the nearest integer. 11, 18,25,...

Zigmanuir [339]

Answer:

8424

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

8/x-5 - 9/x-4 = 5/x^2-9x+20

adelina 88 [10]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2752942

_______________

Solve the equation:

$\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}$

Reduce the fractions at the left side so that they have the same denominator:

$\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}$

Numerators must be equal:

$\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}$

I hope this helps. =)

Tags: rational equation fraction solution algebra

7 0

2 years ago