6^3 = 216. When using chance, take the number of outcomes(six for a die) and raise it to the power of the number of repetitions
The triangle drawn in the question shows a small single line drawn across two sides of the triangle.
This means that those two sides are equal in length.
Hence, the triangle is an isosceles triangle.
In isosceles triangles, the angles opposite to the equal sides are also equal.
Hence, we know that the two angles other than x is 56°.
The sum of the interior angles of a triangle is 180°
x + 56 + 56 = 180
x = 180 - 56 - 56
x = 180 - 112
x = 68°
Hence, the answer is A.
Answer:
A) 113 m²
Step-by-step explanation:
Use the formula for area of a circle.
Area of a circle = πr²
π = pi = 3.14 (rounded)
r = radius
² = the power of 2, or the radius (in this case) multiplied by itself.
Note that the radius is given, and it is 6. Plug in 6 for the radius:
A (circle) = π(6)²
First, solve the power. Multiply 6²:
6² = 6 * 6 = 36
Next, multiply 36 with π (3.14):
36 x 3.14 = 113.04
A) 113 m² is your closest answer.
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Let the angles be A and C
<span>Complimentary angles, therefor </span>
<span>A + C = 90 </span>
<span>The measurement of the complement of an angle exceeds the measure of the angle by 25% (you used a % sign so I'll use % instead of degrees) </span>
<span>C = 1.25A </span>
<span>Substitute </span>
<span>A + 1.25A = 90 </span>
<span>2.25A = 90 </span>
<span>A = 40º <----- </span>
<span>C = 50º <----- </span>
<span>------------------------------ </span>
<span>If you meant 25º </span>
<span>A + (A + 25) = 90 </span>
<span>2A + 25 = 90 </span>
<span>2A = 65 </span>
<span>A = 32.5º </span>
<span>C = 57.5º</span>
Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
