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boyakko [2]
3 years ago
10

3. Angle g and angle h are:

Mathematics
2 answers:
Oduvanchick [21]3 years ago
7 0

Answer: Adjacent angle you can look it up you know that right

mihalych1998 [28]3 years ago
3 0
Answer is A: Adjacent
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Linda scored 35, 47, and 42 points on three tests. How many points should she get on the fourth test to get an average score of
Taya2010 [7]
(35+47+42+x)/4 = 50
(35+47+42+x)=4*50
124 +x  = 200
x=200 - 124
x= 76

Answer B. 76 points.
6 0
4 years ago
FUNCTIONS HELP ME ASAP
deff fn [24]

Hey there! :)

Answer:

f(-2) = 2.

Step-by-step explanation:

To find f(-2), we simply need to look at the y value at x = -2.

At x = -2, the y value is equal to 2.

Therefore:

f(-2) = 2.

3 0
3 years ago
Read 2 more answers
Which situation is not an example of a digital transaction?
Alchen [17]

Answer:

C

Step-by-step explanation:

mailing a check is the mechanical transport of a physical object representing a monetary amount for payment.

that act itself is not electronic.

at the end in today's times the final act to get the money from Elsa'a account into the account of the utility company will be most certainly electronic, but we don't know (it could still end up in the old fashioned way that somebody from the utility company goes with the check to their bank, cash that check in and then use that cash in some way). and Elsa would not be involved. for her the payment was manual, "mechanical", and not electronic.

7 0
2 years ago
Read 2 more answers
A) 30°<br> B) 36°<br> C) 42°<br> D) 55°<br> **Include explanation** <br> To be marked brainliest!
m_a_m_a [10]

Answer:

B, 36 degrees

Step-by-step explanation:

Because AB and CB are congruent, so are arcs AB and CB. Therefore, 3x+12=6x-12

24=3x

x=8

3x+12=3(8)+12=24+12=36 degrees, or answer choice B. Hope this helps!

6 0
3 years ago
An equilateral triangle is inscribed in a circle of radius 6r. Express the area A within the circle but outside the triangle as
Paul [167]

Answer:

A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}

Step-by-step explanation:

We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is \frac{a}{\sqrt{3}}=R.

Upon substituting our given values, we will get:

\frac{5x}{\sqrt{3}}=6r

Let us solve for r.

r=\frac{5x}{6\sqrt{3}}

\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}

We know that area of an equilateral triangle is equal to \frac{\sqrt{3}}{4}s^2, where s represents side length of triangle.

\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2

The area within circle and outside the triangle would be difference of area of circle and triangle as:

A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}

We can make a common denominator as:

A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}

A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}

Therefore, our required expression would be A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}.

7 0
3 years ago
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