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nadezda [96]
3 years ago
9

32 is 64% of what number?

Mathematics
1 answer:
Alex73 [517]3 years ago
8 0
To find percentage, you must divide 32 by the decimal percent. In this case, it is .64
The equation is X/P=Y, with Y being the answer.
When you plug the numbers in, you get this. 32 / .64 = Y
Since 32/ .64 is 50, the correct response is 50. 
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VI. PRACTICE TASKS
dimulka [17.4K]

Step-by-step explanation:

1 _ 1.215

2_ 4.0608

3_1.5225

4_0.2875

5_3.6064

8 0
2 years ago
answer to #10 and #13. can’t figure out how to use interval notation to write domain and range for graphs.
Lyrx [107]
Interval notation is used to write a set of real numbers from one value to another value.
On the left, you start with left parenthesis or left bracket.
Then you follow by two numbers separated by a comma.
You then finish with a right parenthesis or right bracket.
To include a number, use a square bracket.
To exclude a number use parenthesis.
To write the set of numbers, you need to list the smallest number in the set followed by the largest number in the set. An interval is always stated with two numbers, from the smallest in the set to the largest in the set. The numbers are always separated by a comma.

Examples:

1) All numbers from 6 to 10, including 6 and 10.
Algebra: 6 <= x <= 10
Interval: [6, 10]
Notice brackets since both 6 and 10 are included in this interval.

2) All number from 5 to 20, including 5 but not including 20.
Algebra 5 <= x < 20
Interval: [5, 20)
Bracket with 5 means include 5. Parenthesis with 20 means 20 is not included.

3) All numbers greater than or equal to 7.
Algebra: x >= 7
Interval: [7, ∞)
The 7 has a bracket because it is included. Infinity always has parenthesis.
With the infinity symbol, always use parenthesis, not square bracket.

4) All numbers less than -5.
Algebra: x < - 5
Interval: (-∞, 5)

Now for your problems.

10.
This is a line. Both the domain and range all all real numbers.
That means the interval is from negative infinity to positive infinity.
(-∞, ∞)
Both the domain and range are that same interval, all real numbers, from negative infinity to positive infinity.

13.
The domain is all real numbers as you can see the x-coordinates extend left forever and right forever. The domain is the same interval as the domain and range of problem 10.

The range is zero and all positive numbers.
You can think of it a all values of y such that y is greater than or equal to zero. Notice that zero is included in the interval.

[0, ∞)

Since zero is included, we use a left bracket, not left parenthesis.
With infinity, we alyways use parentheses, not brackets.
8 0
3 years ago
The ratio of coloured papers to wight papers is 3:8.If there are 48 wight papers, how many coloured papers are there?
mestny [16]

Answer:  The total number of colored papers is 18.

Step-by-step explanation:

Given: The ratio of coloured papers to wight papers is 3:8.

Let total number of colored papers = 3x

and total number of wight papers = 8x

Since, there are 48 wight papers , then

8x=48

\Rightarrow\ x=6

Now , number of colored papers = 3(6)=18

Hence, the total number of colored papers is 18.

4 0
3 years ago
For what value of a should you solve the system of elimination?
SIZIF [17.4K]
\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}

\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20
\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12

\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}

3a-10y=-8 \ \textgreater \  \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}
3a-10y-3a=-8-3a

\mathrm{Simplify} \ \textgreater \  -10y=-8-3a \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}-10
\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}

Simplify more.

\frac{-10y}{-10} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \  \frac{10y}{10}

\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \  y

-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{-8-3a}{-10}

\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \  -\frac{-3a-8}{10} \ \textgreater \  y=-\frac{-8-3a}{10}

\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \  6x+10\cdot \frac{8}{10-3a}=20

10\cdot \frac{8}{10-3a} \ \textgreater \  \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{8\cdot \:10}{10-3a}
\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \  \frac{80}{10-3a}

6x+\frac{80}{10-3a}=20 \ \textgreater \  \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}
6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}

\mathrm{Simplify} \ \textgreater \  6x=20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \  \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}

\frac{6x}{6} \ \textgreater \  \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \  x

\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \  \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \  \frac{20-\frac{80}{-3a+10}}{6}

20-\frac{80}{10-3a} \ \textgreater \  \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \  \frac{20}{1}-\frac{80}{-3a+10}

\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10

Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \  \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}
\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \  \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \  \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}

20\left(-3a+10\right)-80 \ \textgreater \  Rewrite \ \textgreater \  20+10-3a-4\cdot \:20

\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \  20\left(-3a+10-4\right) \ \textgreater \  Factor\;more

10-3a-4 \ \textgreater \  \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \  -3a+6 \ \textgreater \  Rewrite
-3a+2\cdot \:3

\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \  3\left(-a+2\right) \ \textgreater \  3\cdot \:20\left(-a+2\right) \ \textgreater \  Refine
60\left(-a+2\right)

\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \  \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \  \frac{10\left(-a+2\right)}{\left(-3a+10\right)}

\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \   \frac{10\left(-a+2\right)}{-3a+10}

Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}

Hope this helps!
7 0
3 years ago
Read 2 more answers
How to do algorithm
Rus_ich [418]
Take the 5 in 15 and multiply it by 2. Write it. Then take the 5 and multiply it by 1. Write it. On the next line write a zero and then take the 1 multiplied by 2 and then 1 times 1. Here is an example... it's a little complicated to explain but here...

12
15
----
60
120
-------
180

So sorry I can't explain things well...
6 0
3 years ago
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