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marissa [1.9K]
3 years ago
6

5 x - 3 = 27 , x=___ help

Mathematics
2 answers:
nexus9112 [7]3 years ago
6 0

Answer: x=6

Step-by-step explanation:

Given

5x-3=27

Add 3 on both sides

5x-3+3=27+3

5x=30

Divide 5 on both sides

5x/5=30/5

x=6

Hope this helps!! :)

Please let me know if you have any questions

geniusboy [140]3 years ago
5 0
5x-3=27 +3
5x=30 divide by 5
x= 6
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Nineteen million, two hundred sixty-six thousand, four hundred twenty! :)
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prohojiy [21]

Answer:

x=−5 and y=1

Step-by-step explanation:

7 0
3 years ago
What is the recursive rule for the sequence 2, -1, 1/2 , -1/4
Helga [31]

Answer:

a_n=\frac{-1}{2}a_{n-1}

a_1=2

Step-by-step explanation:

The recursive rule is a term defined in terms of other terms in the sequence.

The is a geometric sequence because it has a common ratio.

The common ratio can be found by dividing a term by previous term.

For example, all of these are equal:

\frac{-1}{2}

\frac{\frac{1}{2}}{-1}

\frac{\frac{-1}{4}}{\frac{1}{2}}

They are all equal to \frac{-1}{2}.

So we are saying:

\frac{\text{term}}{\text{previous term}}}=\frac{-1}{2}

More formally:

\frac{a_n}{a_{n-1}}=\frac{-1}{2}.

Multiply both sides by a_{n-1}:

a_n=\frac{-1}{2}a_{n-1}

When doing recursive form, you need to state a term of the sequence (or more depending on the recursive form you have).

So the first term is 2.

So the full thing for the answer is:

a_n=\frac{-1}{2}a_{n-1}

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8 0
3 years ago
CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
(a-b)(a^2+ab+b^2)=a^3-b^3

In particular,

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

so we have

\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}

For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
Please help me somebody! PLeaseee!
Solnce55 [7]
It should be Expand by distributing terms.
8
×
7
+
8
x
8×7+8x

2 Simplify
8
×
7
8×7 to
5
6
56.

56+8x
ANSWER: 56+8x
3 0
3 years ago
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