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koban [17]
3 years ago
15

Write 0.2 as a fraction

Mathematics
1 answer:
kompoz [17]3 years ago
6 0
As a fraction it would be 2/10 or in its simplest form 1/5
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One number is five less than a second number. Five times the first is 15 more than 6 times the second. Find the numbers
Shalnov [3]

Answer:

x = -45, y = -40

Step-by-step explanation:

Based on the first clue, we find out that

x + 5 = y

5x = 6y + 15

We can re-write the first equation to x = y - 5.

Therefore, the second equation can also be written as 5(y - 5) = 6y + 15

This can be simplified to 5y - 25 = 6y + 15

We can subtract 6y from each side, and add 25 to each side;

5y - 6y = 15 + 25

Therefore, -y = 40

Therefore y = -40.

Now let's plug this back into the original equation, to get : x  + 5 = -40

x=-45

4 0
3 years ago
What is this expression in simplified form 5/2 times 9/6? A.90 B.90/3 C.45/2 D. 45/3
alekssr [168]

5√2 · 9√6

Simplify.

5 × 9 √ 2 x 6 ⇒ Multiply 2 × 6.

5 × 9 √12

Simplify √12 to 2√3.

5 × 9 × 2√3 ⇔ Multiply

90√3

Therefore, the <u>correct alternative</u> is <u>option "B".</u>

8 0
2 years ago
Find the slope of the line that passes through (6, 7) and (2, 10). and simplify if needed thx guys.
tekilochka [14]

Answer:

-\frac{3}{4}

Step-by-step explanation:

the equation for finding the slope of a line when given two points is \frac{y_2-y_1}{x_2-x_1}, aka the change in y over the change in x.

pick one of your coordinate pairs to be y_2\\ and x_2. it doesn't matter which coordinate pair you choose as long as you keep them as y_2\\ and x_2. the remaining coordinate pair will be y_1 and x_1.

for this example, i'll use (2, 10) for y_2\\ and x_2 and (6, 7) for y_1 and x_1.

<em>**before i begin, i just want to note that you can do these next four steps in any order that you want. i personally prefer to plug in my y-values first and then my x-values, but you can choose to instead plug in the values of each coordinate pair (like starting by plugging in the coordinate pair (2, 10) with 10 for </em>y_2\\ and 2 for x_2<em>). it's up to you. i'm going to explain the steps by plugging in my y-values first and then my x-values because that's the way i normally do it.</em>

<em />

first, start by plugging in the y-value from the coordinate pair of your choosing in for y_2\\. since i chose (2, 10) for y_2\\ and x_2, i'll plug in 10 for y_2\\.

\frac{y_2-y_1}{x_2-x_1} ⇒ \frac{10-y_1}{x_2-x_1}

then plug in the remaining coordinate pair's y-value in for y_1. since the coordinate pair that's left is (6, 7), i will plug in 7 for y_1.

\frac{y_2-y_1}{x_2-x_1} ⇒ \frac{10-7}{x_2-x_1}

now i'm going to plug in the x-values. i chose (2, 10) to plug in for y_2\\ and x_2, so now i'll plug in 2 for x_2.

\frac{y_2-y_1}{x_2-x_1} ⇒ \frac{10-7}{2-x_1}

and all that's left to plug in is the x-value from (6, 7), so i will plug that in for x_1.

\frac{y_2-y_1}{x_2-x_1} ⇒ \frac{10-7}{2-6}

after plugging in all the values, you have \frac{10-7}{2-6}.

subtract 10 - 7 as well as 2 - 6.

\frac{10-7}{2-6} ⇒ \frac{3}{-4}

\frac{3}{-4} cannot be simplified, therefore the slope of the line is \frac{3}{-4} or -\frac{3}{4}.

i hope this helps! have a lovely day <3

7 0
3 years ago
Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.
Leya [2.2K]

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6

6 0
3 years ago
GIVING EXTRA POINTS AND BRAINLIEST ANSWER QUICKLY PLEASE
sukhopar [10]

Answer:

j=49

Step-by-step explanation:

3 0
3 years ago
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