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Orlov [11]
3 years ago
9

I need help please?!!!

Mathematics
2 answers:
nydimaria [60]3 years ago
5 0

Answer: 27.5

Step-by-step explanation:

14.5 + 13 = 27.5

loris [4]3 years ago
4 0
The answer should be 27.5 bc u add them together.
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Which of the following could be points on the unit circle?
ipn [44]

Answer:

Points A and C are on the unit circle

Step-by-step explanation:

The unit circle is a circle of radius 1, that is, all the points that are inside the circle have the sum of the squares of its coordinates that are at most 1. Here, we have to test this for each option.

In options B and D, the coordinates 13/7 and 4/3, respectively, means that this sum will be larger than 1, and this points will not be on the unit circle. Now we text for options A and C.

A. ( 5/13, 12/13)

\sqrt{(\frac{5}{13})^2+(\frac{12}{13})^2} = \sqrt{\frac{25}{169}}+\sqrt{\frac{144}{169}} = \sqrt{\frac{169}{169}} = \sqrt{1} = 1

Since the sum of the squares is 1, the point is on the unit circle.

C. (1/3, 2/3)

\sqrt{(\frac{1}{3})^2+(\frac{2}{3})^2} = \sqrt{\frac{1}{9}}+\sqrt{\frac{4}{9}} = \sqrt{\frac{5}{9}}

Since the sum of the squares is less than 1, the point is on the unit circle.

5 0
3 years ago
Will give brainliest
lubasha [3.4K]

Answer:

11

Step-by-step explanation:

7 0
3 years ago
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Three girls share 12 000 in the ratio 2:3:7.Calculate how much each girl will get​
Serjik [45]

Answer:

Step-by-step explanation:

Girl 1:2000 shares

girl 2:3000 shares

girl 3:7000 shares

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4m * 0 * 3m+ 0 name the property
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Find side lengths C and H
Nuetrik [128]

Answer:

\displaystyle c=\frac{2\sqrt{3}}{3}\\\\\displaystyle h=7\sqrt{2}\\

Step-by-step explanation:

<u>Right Triangles</u>

A right triangle can be identified by the fact it has an internal angle of 90°. In a right triangle, the trigonometric ratios stand.

Let's consider the triangle to the left.  We need to calculate side c, one of the legs of the triangle. We can use the angle adjacent to it (60°) or the angle opposite to it (30°) with the appropriate trigonometric ratio.

We'll use the adjacent angle, and

\displaystyle tan60^o=\frac{2}{c}

Solving for c

\displaystyle c=\frac{2}{tan60^o}=\frac{2}{\sqrt{3}}

Rationalizing

\displaystyle c=\frac{2}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}

Now for the triangle to the right. The side h is the hypotenuse. Again, any of the two angles can be used (though they are equal, for it's an isosceles triangle). For any of them it is true that

\displaystyle sin45^o=\frac{7}{h}

Solving for h

\displaystyle h=\frac{7}{sin45^o}=\frac{7}{\frac{\sqrt{2}}{2}}=\frac{14}{\sqrt{2}}

Rationalizing

\displaystyle h=\frac{14}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{14\sqrt{2}}{2}=7\sqrt{2}

3 0
4 years ago
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