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IgorC [24]
2 years ago
11

AC is a tangent to the circle below at point B.

Mathematics
1 answer:
artcher [175]2 years ago
7 0

y=78^{\circ} (alternate interior angles theorem)

x=78^{\circ} (alternate segment theorem)

z=24^{\circ} (angles in a triangle add to 180 degrees)

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If the length of FG is 6 units, what is the length of EH?
QveST [7]

FG goes through the center of the circle, so FG is diameter.

d = 2r \\ r =  \dfrac{d}{2}  \\  \\ r =  \dfrac{6}{2}  = 3

EH=r=3

Answer: EH=3

4 0
3 years ago
Help me with problem 8a and b
svetlana [45]

a. The bridge can hold at most 1000 pounds. This tells you everyone's accumulative weight needs to be less than or equal to 1000. The problem does not give your friend's weight... yet. Let x = your friend's weight.

Your friend's weight (x) plus your weight (156) plus everyone else's weight (675) must be less than or equal to (≤) 1000 pounds.

x + 156 + 675 ≤ 1000

Combine like terms (the constants) to simplify and get your final inequality:

x + 831 ≤ 1000

b. Now, we find out your friend weighs 182 pounds. x = 182. We plug this into the above equation. If it results in a true statement, then you both can walk across the bridge.

182 + 831 ≤ 1000

1013 ≤ 1000, is not a true statement.

No. You both cannot walk across the bridge as is because the weight of the people on the bridge would be 13 pounds over the weight limit.

8 0
3 years ago
And he runs the same number of miles ask every day. His total distance run for one week is less than 60 miles. Which any quality
stiv31 [10]

Answer:

around 8 or 8.5 miles a day for 7 days

Step-by-step explanation:

60 divided 7 equals 8.5

3 0
3 years ago
Which measurement below has four significant figures? A. 14,052 km B. 12.05 L C. 0.0230 kg D. 12.501 m
Dmitrij [34]
The answer is B 12.05L
5 0
4 years ago
Question 7(Multiple Choice Worth 1 points)
lana66690 [7]

Answer:

Option A.

Step-by-step explanation:

The given function is

f(x)=0.15x^2-6x+400

Leading coefficient is positive, so it is an upward parabola. Vertex of an upward parabola is the point of minima.

Vertex of a parabola f(x)=ax^2+bx+c is

Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)

In the given function a=0.15, b=-6 and x=400.

-\dfrac{b}{2a}=-\dfrac{-6}{2(0.15)}=20

At x=20,

f(20)=0.15(20)^2-6(20)+400=340

So, minimum value of f(x) is 340 at x=20.

From the given table it is clear that the minimum value of g(x) is 55 at x=70.

Since 55 < 340, therefore, g(x) has a lower minimum values, i.e, (70,55).

Hence, option A is correct.

4 0
4 years ago
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