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2 years ago

48/50 - 20/25 please Answer this ASAP please

Mathematics

1 answer:

Anuta_ua [19.1K]2 years ago

4 0

Answer:

0.16

Step-by-step explanation:

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Angle $eab$ is a right angle, and $be = 9$ units. what is the number of square units in the sum of the areas of the two squares

Alla [95]

Let the measure of side AB be x, then, the measue of side AE is given by

$AE=\sqrt{9^2-x^2}$ .

Now, ABCD is a square of size x, thus the area of square ABCD is given by

$Area=x^2$

Also, AEFG is a square of size $\sqrt{9^2-x^2}$ , thus, the area of square AEFG is given by

$Area=\left(\sqrt{9^2-x^2}\right)^2=9^2-x^2=81-x^2$

<span>The sum of the areas of the two squares ABCD and AEFG is given by

$x^2+81-x^2=81$

Therefore, </span>the number of square units in the sum of the areas of the two squares <span>ABCD and AEFG is 81 square units.</span>

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2 years ago

If the largest angle of a triangle is 120° and it is included between sides of 1.5 and 0.5, then (to the nearest tenth) the larg

12345 [234]

Answer:

1.8 units.

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other and the given angle is the included angle. This means that the angle is formed by the intersection of the two lines. Therefore, cosine rule will be used to calculate the length of the largest side of the triangle. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that a=0.5, B=120°, and c=1.5. Plugging in the values:

b^2 = 0.5^2 + 1.5^2 - 2(0.5)(1.5)*cos(120°).

Simplifying gives:

b^2 = 3.25.

Taking square root on the both sides gives b = 1.8 (rounded to the nearest tenth).

This means that the length of the third side is 1.8 units!!!

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Convert an annual rate of 0.224% to a generation rate (20 years)

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2.78 that will be the awnser if you go gor it

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2/3(2/8 + 4/8) = 2/3(1/4 + 1/2)
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Which of the following equations have a solution of y = -22 ? Select all that apply.

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Answer:

Where are the equations?

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