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nikklg [1K]
3 years ago
15

What is the slope? (IXL)

Mathematics
1 answer:
Artemon [7]3 years ago
8 0

Answer:

2/4 but simplified 1/2

Step-by-step explanation:

rise over run.

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For every 14 ice cream cones you buy, you get 2 for free.how many cones will you get if you buy 56 cones
lianna [129]
Divide 56/14=4. That means that there are 4 14ns since you will get 2 for every 14 cones. Thay means that you will get 8 cones for free.
6 0
3 years ago
Can anyone please explain? Need some help :)
DedPeter [7]

Answer:

93.5 square units

Step-by-step explanation:

Diameter of the Circle = 12 Units

Therefore:

Radius of the Circle = 12/2 =6 Units

Since the hexagon is regular, it consists of 6 equilateral triangles of side length 6 units.

Area of the Hexagon = 6 X Area of one equilateral triangle

Area of an equilateral triangle of side length s = \dfrac{\sqrt{3} }{4}s^2

Side Length, s=6 Units

\text{Therefore, the area of one equilateral triangle =}\dfrac{\sqrt{3} }{4}\times 6^2\\\\=9\sqrt{3} $ square units

Area of the Hexagon

= 6 X 9\sqrt{3} \\=93.5$ square units (to the nearest tenth)

7 0
3 years ago
This figure shows circle O with inscribed ∠XYZ. m∠XYZ=76∘ What is the measure of XYZ?
Cloud [144]

Answer:

152

Step-by-step explanation:

Since the measure of angle XYZ is 76 degrees, we know that the measure of XYZ would be twice the amount of the angle measurement due to the square root theorem.  In this case 76 times 2 or 76 plus 76 equals 152 degrees.  152 degrees is your final answer.

6 0
3 years ago
Read 2 more answers
HELP PLS’ Suppose you roll a die. Find the probability of the given event. Simplify your answer.
NikAS [45]

The answer is 2) 2/3

7 0
3 years ago
Find the complex fourth roots \[-\sqrt{3}+\iota \] in polar form.
____ [38]

Let z=-\sqrt3+i. Then

|z|=\sqrt{(-\sqrt3)^2+1^2}=2

z lies in the second quadrant, so

\arg z=\pi+\tan^{-1}\left(-\dfrac1{\sqrt3}\right)=\dfrac{5\pi}6

So we have

z=2e^{i5\pi/6}

and the fourth roots of z are

2^{1/4}e^{i(5\pi/6+k\pi)/4}

where k\in\{0,1,2,3\}. In particular, they are

2^{1/4}e^{i(5\pi/6)/4}=2^{1/4}e^{i5\pi/24}

2^{1/4}e^{i(5\pi/6+2\pi)/4}=2^{1/4}e^{i17\pi/24}

2^{1/4}e^{i(5\pi/6+4\pi)/4}=2^{1/4}e^{i29\pi/24}

2^{1/4}e^{i(5\pi/6+6\pi)/4}=2^{1/4}e^{i41\pi/24}

7 0
3 years ago
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