Please help me w/ this geometry question.

image attached.

Answers

Answer 1

Answer:
$The \: answer \: is \: 46 °$

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So a 4 digit number is represented ABCD, they are not multiplied together, each letter represents 1 single , whole numberif ABCD times D=DCBAsolve for the values of A and B and C and D with full workings and solution (this problem is from my 5th grade math book, just for fun)

(1+5/6)(1-5/6)

I don't understand it

Answers

(a+b)(a-b) = a^2 - b^2

(1+5/6) (1-5/6) = 1^2 - (5/6)^2 = 1 - 25/36 = 36/36 - 25/36 = (36-25)/36 = 11/36

11/36 is your answer

A triangle is placed in a semicircle with a radius of 9 cmFind the area of the shaded region. Use the value 3.14 for π
, and do not round your answer. Be sure to include the correct unit in your answer.

Answers

Considering that the shaded area is the part of the semi circle outside the triangle:
Area of circle= Pi x radius2 (this means radius squared)
Area of triangle= 1/2 x base x height
1/2 (because it is a semi circle) x 3.14 (in replacement of pi as directed by question) x 9squared= 127.17

127.17 - (1/2 x 18 x 9) = 46.17cmsquared

Final answer:

The area of the shaded region can be found by subtracting the area of the triangle from the area of the semicircle.

Explanation:

To find the area of the shaded region, we need to calculate the area of the semicircle and subtract the area of the triangle from it. First, let's find the area of the semicircle. The formula to find the area of a semicircle is A = (π * $r^2$ ) / 2, where π is approximately 3.14 and r is the radius. Plugging in the values, we have A = (3.14 * $9^2$ ) / 2 = 127.26 square cm.

Next, let's find the area of the triangle. The formula to find the area of a triangle is A = (base * height) / 2. In this case, the base of the triangle is the diameter of the semicircle, which is 2 * 9 = 18 cm. The height of the triangle is the radius of the semicircle, which is 9 cm. Plugging in the values, we have A = (18 * 9) / 2 = 81 square cm.

Finally, to find the area of the shaded region, we subtract the area of the triangle from the area of the semicircle: 127.26 - 81 = 46.26 square cm. So, the area of the shaded region is 46.26 square cm.

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-\dfrac{1}{6} \times \left(-\dfrac{9}{7}\right)− 6 1 ×(− 7 9 )minus, start fraction, 1, divided by, 6, end fraction, times, left parenthesis, minus, start fraction, 9, divided by, 7, end fraction, right parenthesis

Answers

Answer:

$(3)/(14)$

Step-by-step explanation:

The even number of minus signs means the product will be positive. A factor of 3 can be removed to simplify the product. As usual, the numerator of the product is the product of numerators, and the denominator of the product is the product of denominators.

$-(1)/(6) * \left(-(9)/(7)\right)=(1\cdot 9)/(6\cdot 7)=(3\cdot 3)/(3\cdot 14)=\boxed{(3)/(14)}$

The answer to the expression $-(1)/(6) * \left(-(9)/(7)\right) - (1)/(6) * \left(-(9)/(7)\right)$ is 0.

In this expression, we have two terms that are identical: $-(1)/(6) * \left(-(9)/(7)\right)\) and \(-(1)/(6) * \left(-(9)/(7)\right)$ .

When we multiply a negative number by another negative number, the result is positive.

So, both of these terms become positive, and since they are identical, when we subtract one from the other, we get zero.

Now, let's provide a more detailed explanation.

When we multiply $-(1)/(6)\) by \(-(9)/(7)$ , we first multiply the numerators to get 9 and the denominators to get 42.

So, $-(1)/(6) * \left(-(9)/(7)\right) = (9)/(42)$ .

Now, we can simplify $(9)/(42)$ by dividing both the numerator and denominator by their greatest common divisor, which is 3.

So, $(9)/(42)$ simplifies to $(3)/(14)$ .

Therefore, the expression becomes $(3)/(14) - (3)/(14)$ , and when we subtract these two identical fractions, the result is indeed 0.

for such more questions on expression

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twenty-five students use 120 sheets of paper. find each unit rate round to the nearest hundreth if neccasery

Answers

$(25\ students)/(120\ sheets\ of\ paper)=(25)/(120)\ students/sheets\ of\ paper\n\n=(5)/(24)\ students/sheets\ of\ paper\approx0.21\ students/sheet\ of\ paper$