All categories

2 years ago

One number is seven more than a second number. Twice the first is 2 less than 3 times the second. Find the numbers.

Mathematics

1 answer:

Keith_Richards [23]2 years ago

4 0

Step-by-step explanation:

Linear equations. Make 2 new equations from the question. Name the first number as x and the second as y...

You might be interested in

*math please help brainliest*

valentinak56 [21]

Answer:

$\dfrac{3x+4}{2x-2}$

Step-by-step explanation:

Invert and multiply the way you would any fraction division problem. Then cancel common factors from numerator and denominator.

$\displaystyle\frac{\left(\frac{4x^2+2x}{x^2+x-2}\right)}{\left(\frac{8x^2+4x}{3x^2+10x+8}\right)}=\frac{4x^2+2x}{x^2+x-2}\cdot\frac{3x^2+10x+8}{8x^2+4x}\\\\=\frac{2x(2x+1)(3x+4)(x+2)}{4x(2x+1)(x-1)(x+2)}=\frac{3x+4}{2(x-1)}\\\\=\frac{3x+4}{2x-2}$

4 0

3 years ago

What is 04.151515... into a fraction

Sergio039 [100]

Answer:

$\displaystyle 4.\overline{15} = \frac{137}{33}$ .

Step-by-step explanation:

Start by separating this decimal number into its integer part and its fraction part:

$4.151515\cdots = 4 + 0.151515\cdots$

The most challenging task here is to express $0.151515\cdots$ as a proper fraction. Once that fraction is found, expressing the original number $4.151515\cdots$ will be as simple as rewriting a mixed number as an improper fraction.

Let $x = 0.151515\cdots$ . $(x + 4)$ would then represent the original number.

Note that the repeating digits appear in groups of two. Therefore, if the digits in $x$ are shifted to the left by two places, the repeating part will continue to match:

$\begin{aligned}x = 0.&151515\cdots && \\ 100\, x = 15.& 151515\cdots \end{aligned}$ .

Note, that this "shifting" is as simple as multiplying the initial number by $100$ (same as $10$ raised to the power of the number of digits that needs to be shifted.)

Subtract the original number from the shifted number to eliminate the fraction part completely:

$\begin{aligned}&(100\, x) - x \\ &= 15.151515\cdots\\ & \phantom{=}- 0.151515\cdots\\&=15 \end{aligned}$ .

In other words:

$99\, x = 15$ .

$\displaystyle x = \frac{15}{99} = \frac{5}{33}$ .

Therefore, the original number would be:

$\displaystyle x + 4 = \frac{5}{33} = \frac{132 + 5}{33} = \frac{137}{33}$ .

4 0

3 years ago

Laws Of Indicies<br>Can someone please help me with this?

nasty-shy [4]

Answer:

See below

Step-by-step explanation:

$1) : {a}^{5} {b}^{4} {c}^{3} \div {a}^{2} {b}^{3} c \\ \\ = {a}^{5 - 2} {b}^{4 - 3} {c}^{3 - 1} \\ \\ = {a}^{3} {b}^{1} {c}^{2} \\ \\ = {a}^{3} {b} {c}^{2} \\ \\ \\2)\: (x^2 y^3)^3\\\\= x^{2\times 3}y^{3\times 3}\\\|= x^{6}y^{9}\\\\\\3)\:5 {a}^{4} \div ( {a}^{2} \times 3a) \\ \\ = 5 {a}^{4} \div (3 {a}^{2 + 1} ) \\ \\ = 5 {a}^{4} \div (3 {a}^{3} ) \\ \\ = \frac{5}{3} {a}^{4 - 3} \\ \\ = \frac{5}{3} {a} \\ \\ \\ 4)\: \frac{14 {a}^{5} }{2 {a}^{3} \times 7 {a}^{4} } \\ \\ = \frac{14 {a}^{5} }{2 \times 7 {a}^{3 + 4} } \\ \\ = \frac{ \cancel{14} {a}^{5} }{ \cancel{14}{a}^{3 + 4} } \\ \\ = \frac{ {a}^{5} }{ {a}^{7} } \\ \\ = {a}^{5 - 7} \\ \\ = {a}^{ - 2} \\ \\ = \frac{1}{{a}^{2} } \\ \\$