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3 years ago

Identify each figure in as many ways as possible

1 answer:

5 0

Parallelogram
- quadrilateral (four sided figure)
- oposite sides parallel
Rectangle
- quadrilateral
- equiangular
- right angle (90° angle)
Isosceles Trapezoid
- quadrilateral
- pair of parallel sides
- called "trampezium" in the UK
Square
- regular quadrilateral
- equiangular
- right angles (90°)

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Solve the formula V=h for r.

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Answer:

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Step-by-step explanation:

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3 years ago

Please help !! Simplify the following expression

puteri [66]

8x^4 + x^3 - 4x^2 +1. I believe the answer is B. Please correct me if I am wrong.

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3 years ago

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Which of these equations does not have any solutions?

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the answer is C........

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3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

A golf tournament has five rounds. The players are given scores for each round based on how many strokes they are above or below

frozen [14]

We have the following table:

Round Number of Strokes Above or Below 72

1 -5

2 +6

3 -2

4 -4

5 +5

So, the expression that represents the total number of strokes is:

72 + (Number of strokes Above or below 72)

Therefore, for every round we get:

Round Number of Strokes Above or Below 72 Total number of strokes

1 -5 72 - 5 = 67

2 +6 72 + 6 = 78

3 -2 72 - 2 = 70

4 -4 72 - 4 = 68

5 +5 72 + 5 = 77

The total number of strokes is:

67 + 78 + 70 + 68 + 77 = 360

Additionally, the average is the total number of strokes divided by the number of rounds, so it is equal to:

$\frac{360}{5}=72$

Answer: Total: 360 Strokes

Average: 72 Strokes

6 0

1 year ago