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2 years ago

What is the answer to: Find the first three terms: nth term=3n+4?

Mathematics

2 answers:

11111nata11111 [884]2 years ago

6 0

Answer:

7, 10, 13

Step-by-step explanation:

1) n = 1: 3(1) + 4 = 3 + 4 = 7

2) n = 2: 3(2) + 4 = 6 + 4 = 10

3) n = 3: 3(3) + 4 = 9 + 4 = 13

kramer2 years ago

5 0

Answer:

The Three terms are 3, n, and four.

Step-by-step explanation:

Terms are number are variable, so the terms are 3, n, and 4.

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3 years ago

One month, ruby worked 8 hours more than isaac, and svetlana worked 4 times as many hours as ruby. together they worked 136 hour

Travka [436]

Let Issac work for x hours, and Ruby works 8 hours more than of Issac, so Ruby work for $(x+8)$ hours.

And Svetlana worked four times of Ruby, therefore Svetlana worked for

$4(x+8) hours$

And total hours they worked together is 136. That is

$x+x+8 + 4(x+8) = 136
\\
x+x+8 +4x+32=136
\\
6x+40 =136
\\
6x=96
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6 0

3 years ago

If five times the successor of a number is added to an original number the sum is 83

Alexus [3.1K]

Answer:

Let the original number be x

Successor is defined as the number which comes immediately after a particular number.

also, the successor of a whole number is the number obtained by adding 1 to that number.

Then, the successor of a number x is, x+1

As per the given condition :

we have;

$5(x+1)+x =83$

Using distributive property on LHS (i.e, $a\cdot(b+c)=a\cdot b+a\cdot c$ )

Then, we have

5x+5+x=83

Combine like terms;

6x+5=83

Subtract 5 from both the sides we get;

6x+5-5=83-5

Simplify:

6x=78

Divide both side by 6,

$\frac{6x}{6} =\frac{78}{6}$

Simplify:

x =13

Therefore, the original number x is, 13

8 0

2 years ago

P³ = 1/8 please help me if anybody can .

ivanzaharov [21]

Answer:

$p= \dfrac{1}{2}$

Step-by-step explanation:

Given equation:

$p^3=\dfrac{1}{8}$

Cube root both sides:

$\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}$

$\implies p= \sqrt[3]{\dfrac{1}{8}}$

$\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:$

$\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}$

$\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:$

$\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}$

$\textsf{Apply exponent rule} \quad 1^a=1:$

$\implies p= \dfrac{1}{8^{\frac{1}{3}}}$

Rewrite 8 as 2³:

$\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}$

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

$\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}$

Simplify:

$\implies p= \dfrac{1}{2^{\frac{3}{3}}}$

$\implies p= \dfrac{1}{2^{1}}$

$\implies p= \dfrac{1}{2}$

3 0

10 months ago

Read 2 more answers