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Anestetic [448]
3 years ago
10

What is the area of a circle with radius 5 cm

Mathematics
2 answers:
bija089 [108]3 years ago
6 0

Answer:

A≈78.54cm²

Step-by-step explanation:

aalyn [17]3 years ago
5 0

Answer:

78.55cm^2

Step-by-step explanation:

Area of a circle is given as πr^2

That's is ,

A = πr^2

Let's use π = 3.143

r = 5cm

A= 3.142 * 5^2

A= 3.142 * 25

A = 78.55 cm^2

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Answer:

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Step-by-step explanation:

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3 0
2 years ago
A student inscribes a triangle within a semicircle. What is the measure of ∠XYZ?
nata0808 [166]

Answer: a. 90°


Step-by-step explanation:

We know that the in a circle, the measure of an inscribed angle is half the measure of the central angle with the same intercepted arc.

In the problem∠XYZ is the inscribed angle

∠XYZ=\frac{1}{2}\ arc(XZ)

⇒ ∠XYZ=\frac{1}{2}\angle{XZ}

Since XZ is a diameter of the circle which is a line segment, thus ∠XZ=180°

∴ ∠XYZ=\frac{1}{2}180^{\circ}=90^{\circ}

∴ ∠XYZ=90^{\circ}

Therefore, a. 90° is the measure of ∠XYZ.


6 0
3 years ago
Read 2 more answers
Find the value of x to the nearest tenth​
xz_007 [3.2K]

Answer:

9.0

Step-by-step explanation:

x = 7 / cos 39°

= 7 / 0.7771

= 9.0

4 0
3 years ago
Which function is the inverse of function f? f(x)=9x^2-12
Allisa [31]

Answer:

f−1(x)=√x+123,−√x+123

Step-by-step explanation:

5 0
3 years ago
the angle theta is in the second quadrant and cos theta = -2/√29 determine possible coordinates for point P on the terminal arm
Gekata [30.6K]

\cos(\theta)=-\frac{2}{\sqrt{29}} and $\theta$ lies in $2^{\text{th}}$ quadrant.

where, $x-$ coordinate is negative, and $y-$ coordinate is positive

so it can't a.

now, cosine means, side adjacent over the hypotenuse, in Cartesian plane, that will be $x-$ coordinate over the distance from origin.

Assume the triangle , with base $2$ units and hypotenuse $\sqrt{29}$ and it's in second quadrant. (so \cos(\theta)=-\frac{2}{\sqrt{29}})

now, the leftmost point on $x-$ axis is , obviously $(-2,0)$

and by Pythagoras theorem, we can find the perpendicular side, that will be $y^2=(\sqrt{29})^2-(2)^2\implies y=5$

so the coordinates of the upper vertex is $(-2,5)$, each point lying on this "ray" should have equal ratio of respective coordinates. i.e. $\frac25=\left|\frac xy\right| $

and it should lie on second quadrant, so $x<0 \, y>0$

Option d satisfies this.

6 0
3 years ago
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