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3 years ago

5

Three consecutive even integers such that 12 less than three times the largest is equal to 96 more than the sum of the first and

second. what is the first even integer?

1 answer:

Marat540 [252]3 years ago

8 0

<span>Let n, n+2, and n+4 represent the three consecutive even integers.

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Solve for B.<br><br> A=3B+7C

Archy [21]

Hey there!!

Given:

... A=3B+7C

Subtracting 7C on both sides:

... A-7C=3B

Dividing by B on both sides:

... (A-7C)/3=B

<em>Hope my answer helps!</em>

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3 years ago

If 23% of "A" is 46 , then find "A"

neonofarm [45]

Step-by-step explanation:

23percent *A=46

23/100*A=46

A=46*100/23

A=200

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3 years ago

Read 2 more answers

Let: f(x)=x+1 and g(x)=x^2-2 <br><br> find: g(x) - f(x)

weeeeeb [17]

Answer:

x^2-x-3

Step-by-step explanation:

x^2-2-x-1

x^2-x-3

3 0

3 years ago

What’s the answer to this?

Hoochie [10]

Answer:

The missing term is:

$a_2=14$

Step-by-step explanation:

Considering the geometric sequence

$2,\:?,\:98,\:...$

Here,

$a_1=2$

$a_3=98$

$a_2=?$

The general term of a geometric sequence is given by the formula:

$a_n=a_1\cdot \:r^{n-1}$

where $a_1$ is the initial term and $r$ the common ratio.

as

$a_3=a_1\cdot \:r^{3-1}$

$98=2\cdot \:r^{3-1}$ ∵ $a_1=2$

$\mathrm{Switch\:sides}$

$2r^{3-1}=98$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\frac{2r^{3-1}}{2}=\frac{98}{2}$

$r^{3-1}=49$

$r^2=49$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$r=\sqrt{49},\:r=-\sqrt{49}$

$r=7,\:r=-7$

Taking positive value of $r = 7$

$a_2=a_1\cdot \:r^{n-1}$

putting $r = 7$

$a_2=2\cdot \:7^{2-1}$

$a_2=2\cdot \:7$

$a_2=14$

So, the sequence becomes $2,\:14,\:98,\:...$

Therefore, the missing term is:

$a_2=14$

4 0

3 years ago