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3 years ago

How do I solve this equation? I need a refresher on Algebra.

Mathematics

1 answer:

yaroslaw [1]3 years ago

5 0

These equations seem tricky, but aren't as hard as you think. Just solve it like you are regularly adding fractions.

Your common denominator is 9x + 9, so

1/(9x+9) = 36/(9x + 9) - (5x + 5)/9

1 - 36 + 5x + 5 / 9x - 9
30x - 30/9x - 9
10x-10/3x - 3
10(x-1)/3(x-1)
10/3

Hope this helped!

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Ons: Practice

Sedaia [141]

Answer:

$x = 18$

Step-by-step explanation:

Given

$\frac{x}{6} - 7 = -4$

Required

Solve for x

$\frac{x}{6} - 7 = -4$

Add 7 to both sides

$\frac{x}{6} - 7+7 = -4+7$

$\frac{x}{6} = -4+7$

$\frac{x}{6} = 3$

Multiply through by 6

$6 * \frac{x}{6} = 3 * 6$

$x = 3 * 6$

$x = 18$

7 0

3 years ago

Find the common ratio for the geometric sequence defined by the formula: an=40(2‾√)n−1 a n = 40 ( 2 ) n − 1

WITCHER [35]

The ratio of the geometric sequence 40 $2^{n-1}$ is 2.

Given that geometric sequence is 40* $2^{n-1}$ and we have to find the common ratio of all the terms.

Geometric sequence is a sequence in which all the terms have a common ratio.

Nth termof a GP is a $r^{n-1}$ in which a is first term and r is common ratio.

Geometric sequence=40* $2^{n-1}$

We have to first find the first term, second term and third term of a geometric progression.

First term=40* $2^{1-1}$

=40* $2^{0}$

=40*1

=40

Second term=40* $2^{2-1}$

=40* $2^{1}$

=40*2

=80

Third term=40* $2^{3-1}$

=40* $2^{2}$

=40*4

=160

Ratio of first two terms=80/40=2

Ratio of next two terms=160/80=2

Hence the common ratio of geometric sequence is 2.

Learn more about geometric progression at brainly.com/question/12006112

#SPJ1

4 0

2 years ago

Let k ( x ) = 2 x + 3 and m ( x ) = 3 x 2 + 7 x − 10 . Find the indicated value. M ( k ( − 4 ) ) =

murzikaleks [220]

Answer: $30$

Step-by-step explanation:

Given

$k(x)=2x+3$

$M(x)=3x^2+7x-10$

$M\left(k(x)\right)$ is given by replacing the $x$ by $k(x)$ from the $M(x)$

$\Rightarrow M\left(k(x)\right)=3(2x+3)^2+7(2x+3)-10\\\Rightarrow M\left(k(x)\right)=3(4x^2+9+12x)+14x+21-10\\\Rightarrow M\left(k(x)\right)=12x^2+50x+38$