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3 years ago

True of false a circle could be circumscribed about the quadrilateral below?

Mathematics

2 answers:

Volgvan3 years ago

5 0

Answer: True

The reason why this is true is because the opposite angles add up to 180 degrees
120+60 = 180
90+90 = 180
With any inscribed quadrilateral in a circle, the opposite angles must be supplementary (i.e. add to 180 degrees)

Alex3 years ago

4 0

Answer:True

Step-by-step explanation:

Two sides have to equal 180

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C.) 27 and a half. 12 minutes times 5 equals an hour so 5 and a half pages times 5 equals 27 and a half pages per hour.

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3 years ago

Consider equation of circle x2-2x+y2-4y-4=0,if line 2x-y+a=0 is its diameter then find value of a

g100num [7]

Answer:

Step-by-step explanation:

x² - 2x + y² - 4y - 4 = 0

x² - 2x + y² - 4y = 4

complete the squares

(x² - 2x + x₀) + (y² - 4y + y₀) = 4

the unknown number will be the square of half of the x¹ coefficient

(x² - 2x + 1) + (y² - 4y + 4) = 4 + 1 + 4

(x - 1)² + (y - 2)² = 9

This tells us that the circle is centered on (1, 2)

Which is also a point on the diameter

2x - y + a = 0

a = y - 2x

a = 2 - 2(1)

a = 0

4 0

3 years ago

The falcons basketball team beat the tigers by 7 points in a game. The total points scored by both teams was 83. Write and solve

Blizzard [7]

Falcons = 45
Tigers = 38

83-7=76
76/2=38 (Tigers)
38+7=45 ( Falcons)

6 0

3 years ago

Translate this algebraic expression problem 5 take away from d

Xelga [282]

Answer:

5-d

Step-by-step explanation:

7 0

3 years ago

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

KIM [24]

Answer:

The answer is

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$tan(\theta)=-1$

Step-by-step explanation:

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$sin^{2}(\theta)+cos^{2}(\theta)=1$

In this problem we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

$\frac{3\pi}{2}$

The angle $\theta$ belong to the third or fourth quadrant

The value of $sin(\theta)$ is negative

Step 1

Find the value of $sin(\theta)$

Remember

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(\theta)=1-\frac{1}{2}$

$sin^{2}(\theta)=\frac{1}{2}$

$sin(\theta)=-\frac{\sqrt{2}}{2}$ ------> remember that the value is negative

Step 2

Find the value of $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$tan(\theta)=-1$

8 0

3 years ago

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