How to solve the question

Mathematics

2 answers:

Pani-rosa [81]3 years ago

4 0

It should be 90cm
So it says 20cm multiply that by2 which is 40
Then add all the 10.
That is it!

Norma-Jean [14]3 years ago

4 0

Find area of quadrant with radius 10 cm and subtract the triangle. Multiply that by 2 for area at right corner. Let’s call this K
Find area of quadrant with radius 20 cm and subtract a semicircle and square with area of 100 cm^2. Add K to get answer.

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The figure below shows a shaded circular region inside a larger circle: A shaded circle is shown inside another larger circle. T

Taya2010 [7]

Area of larger circle: 3.14 x 5^2 = 78.5

Area of small circle: 3.14 x 2^2 = 12.56

Difference of the 2 areas: 78.5 - 12.56 = 65.94

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3 0

3 years ago

Read 2 more answers

Maximize the objective function P = 2x + 1.5y for the feasible region shown. State the maximum value for P and the ordered pair

umka2103 [35]

Incomplete question. However, let's assume this are feasible regions to consider:

Points:

- (0, 100)

- (0, 125)

- (0, 325)

- (1, 200)

Answer:

<u>Maximum value occurs at 325 at the point (0, 325)</u>

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

Remember, we substitute the points value for x, y in the objective function P = 2x + 1.5y.

- For point (0, 100): P= 2(0) + 1.5 (100) =150

- For point (0, 125): P= 2(0) + 1.5 (125) =187.5

For point (0, 325): P= 2(0) + 1.5 (325) = 487.5

For point (1, 200): P= 2(1) + 1.5 (200) = 302

Therefore, we could notice from the above solutions that at point (0,325) we attain the maximum value of P.

7 0

2 years ago

Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If

natima [27]

Answer:

d) one solution; (4, 1)

Step-by-step explanation:

It often works well to follow problem directions. A graph is attached, showing the one solution to be (4, 1).

_____

You know there will be one solution because the lines have different slopes. Each is in the form ...

y = mx + b

where m is the slope and b is the y-intercept.

The first line has slope -1 and y-intercept +5; the second line has slope 1 and y-intercept -3. The slope is the number of units of "rise" for each unit of "run", so it can be convenient to graph these by starting at the y-intercept and plotting points with those rise and run from the point you know.