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3 years ago

Solve 3x +2=5.

Mathematics

1 answer:

noname [10]3 years ago

5 0

Answer:

x = 1

Step-by-step explanation:

3x + 2 = 5

Inverse oeprations

3x + 2 = 5

-2 -2

3x = 3

/3 /3

x = 1

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What is x?????????????

Mekhanik [1.2K]

Answer:

BGC=180-90

=90

BGC=BGE+EGC

BGC=BGE+15

90=BGE+15

BGE=90-15

BGE=75

BGE=AGF

AGF=X,

X =75

8 0

2 years ago

20 POINTS PLEASE HELP Determine whether each figure has rotational symmetry, line symmetry, both, or neither.

alexgriva [62]

1-rotational
2-rotational
3-rotational
4-neither

(I’m like 99% sure)

4 0

2 years ago

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Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

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

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

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

2 years ago

What is 9x6+(5x7+4x3)63x8

irina [24]

Solution:

$=9*6+(5*7+4*3)63*8\\=54+(35+12)504\\=54+47*504\\=54+23688\\=23742\\$

Answer:

$23742$

Hope this was helpful.

8 0

2 years ago

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So I have to compute the following but i don't know what they mean any help?

Gelneren [198K]

Answer:

Step-by-step explanation:

This is a Combination (as in permutation vs combination) question the symbol (n r) refers to "n choose r". This is sometimes written as nCr

i.e the question is asking you to find how many combinations each will yield when you chose r items from n item without repetition and order does not matter.

I will only do the first question for you and you can just follow the same steps to solve the rest of the questions.

Recall that

$nCr=\frac{n!}{(r!)(n-r)!}$

Consider question a) we are given (5 1) or ₅C₁

we can see that n = 5 and r = 1

If we substitute this into the formula:

₅C₁ = (5!) / [ (1!)(5- 1)!]

= (5!) / [ (5- 1)!]

= (5!) / (4!)

= (5·4·3·2·1) / (4·3·2·1)

= 5

hence ₅C₁ = 5

7 0

2 years ago

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