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Marat540 [252]
3 years ago
12

What percent is 237 of 845

Mathematics
2 answers:
vladimir2022 [97]3 years ago
8 0
1. We assume, that the number 845 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 100% equals 845, so we can write it down as 100%=845.
4. We know, that x% equals 237 of the output value, so we can write it down as x%=237.
5. Now we have two simple equations:
1) 100%=845
2) x%=237
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
100%/x%=845/237
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for 237 is what percent of 845

100%/x%=845/237
(100/x)*x=(845/237)*x       - we multiply both sides of the equation by x
100=3.5654008438819*x       - we divide both sides of the equation by (3.5654008438819) to get x
100/3.5654008438819=x
28.047337278107=x
x=28.047337278107

now we have:
<span>237 is 28.047337278107% of 845</span>
Rzqust [24]3 years ago
4 0
237 is 28,05% of 845     
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determine the value of x for which r parallels s. then find the measure of angle 1 and the measure of angle 2measure angle 1= 50
Maslowich

According to the picture, angles 1 and 2 have the same measure because they are corresponding angles. Make equal both expressions for 1 and 2 and find x.

\begin{gathered} 50-2x=110-42x \\ 42x-2x=110-50 \\ 40x=60 \\ x=\frac{60}{40} \\ x=\frac{3}{2} \end{gathered}

x is 3/2. Replace this value in both expression to find the measure of each angle.

\begin{gathered} \measuredangle1=50-2x \\ \measuredangle1=50-2(\frac{3}{2}) \\ \measuredangle1=50-3 \\ \measuredangle1=47 \\ \measuredangle2=110-42x \\ \measuredangle2=110-42(\frac{3}{2}) \\ \measuredangle2=110-63 \\ \measuredangle2=47 \end{gathered}

The measure of these angles is 47°

7 0
1 year ago
What is the distance between (4,2) and (-1,-3)
Sati [7]
<h2>Solving for the Distance between two Points</h2><h3>Answer:</h3>

5 \sqrt{2} or 7.071

<h3>Step-by-step explanation:</h3>

<em>Please refer to my answer from this Question to know more about distances between two points: <u>brainly.com/question/24629826</u></em>

Given:

(4,2)

(-1,-3)

Solving for the Distance:

\sqrt{(4-(-1))^2 +(2 -(-3))^2} \\ \sqrt{(4 +1)^2 +(2 +3)^2} \\ \sqrt{5^2 +5^2} \\ \sqrt{25 +25} \\ \sqrt{50} \\ \sqrt{25 \cdot 2} \\ 5 \sqrt{2}

<u>Note:</u>

5 \sqrt{2} \approx 7.071

4 0
3 years ago
14 more than 6 times a number is<br> greater than or equal to 28
Studentka2010 [4]

Answer:

6x + 14 ≥ 28

Step-by-step explanation:

6 times a number (6x)

14 more than (so, add 14 to 6x)

greater than or equal to, so this symbol ≥

= 6x + 14 ≥ 28

SOLVED:

6x≥14

x≥2.333333333333333

4 0
3 years ago
A flying carpet flies 2.4 miles with the wind in the same amount of time it flies 1.4 miles against the wind. The wind speed is
Ainat [17]

Answer:

15.2~mph

Step-by-step explanation:

Let the time flown by the flying carpet be t. Then, we have that (2.4)/t=(1.4)/t+2*4.

We multiply both sides of the equation by t to get 2.4=1.4+8t.

We subtract 1.4 from both sides to get 1=8t.

We divide both sides of the equation by t=1/8.

We know that the speed of the flying carpet in still wind is the average of the rates of the speed of the flying carpet with the wind and against the wind.

The speed with wind is (2.4)/(1/8)=19.2mph.

The speed against wind is (1.4)/(1/8)=11.2mph.

The speed in still wind is (19.2+11.2)/2=(30.4)/2=15.2.

Therefore, the answer is \boxed{15.2~mph} and we're done!

4 0
3 years ago
The tables represent two linear functions in a system,
Alenkasestr [34]

Answer:

<h2>(8, -22)</h2>

Step-by-step explanation:

The slope-intercept form of an equation of a line:

y=mx+b

m - slope

b - y-intercept

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

First table:

(-4, 26), (0, 10) → b = 10

m=\dfrac{10-26}{0-(-4)}=\dfrac{-16}{4}=-4

\boxed{y=-4x+10}

Second table:

(-4, 14), (0, 2) → b = 2

m=\dfrac{2-14}{0-(-4)}=\dfrac{-12}{4}=-3

\boxed{y=-3x+2}

We have the system of equations:

\left\{\begin{array}{ccc}y=-4x+10&(1)\\y=-3x+2&(2)\end{array}\right\\\\\text{Put (1) to (2):}\\\\-4x+10=-3x+2\qquad\text{subtract 10 from both sides}\\-4x=-3x-8\qquad\text{add 3x to both sides}\\-x=-8\qquad\text{change the signs}\\x=8\\\\\text{Put the value of x to (2):}\\\\y=-3(8)+2\\y=-24+2\\y=-22

5 0
3 years ago
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