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Scorpion4ik [409]
3 years ago
15

Solve for: a,b, and c

Mathematics
1 answer:
andrew11 [14]3 years ago
3 0

a=2

b=6

c=-6

You can use the substitution method to solve.

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Help me with this problem pls
Vika [28.1K]

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72 degrees

Step-by-step explanation:

360/5=72

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4 years ago
Which shows the expression below simplified?
salantis [7]

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D

Step-by-step explanation:

(7 × 10-7) × 0.06

7 × 0.06 × 10^-7

0.42 × 10^-7

4.2 × 10^-8

7 0
4 years ago
Please help, thank you!! Will mark brainliest!!
muminat

Answer:

(-x,y) To (x,-y)

Step-by-step explanation:

6 0
3 years ago
√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1
Snowcat [4.5K]

Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}


\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s

7 0
3 years ago
Given the lengths of two sides of a triangle, find the range for the length of the third side. Write an inequality.
djverab [1.8K]

Answer:

| a - b | < length of third side < a + b

Step-by-step explanation:

Visualize the two given sides of the triangle (let's call then a and b), joined at the vertex of the triangle, and forming an angle. We can join the other free end of these two segments, with another segment whose length would vary according to how tiny or large the angle is. We can spread the aperture of the angle they form as much as we can just below  (not reaching this angle measure, because in such case, there will be no triangle of tangible area. In such case, the length of the joining segment will be limited by the addition of the two sides:

length of third side < a + b

In the case the aperture of the angle formed by the two given sides is diminished as much as possible to still form a measurable triangle, the angle has to be just larger than zero, and in such case, the segment joining the other to ends of a and b would be just larger than the absolute value of the difference between a and b:

length third side > | a - b|

These are the two extreme cases, and the length of the third side must be within these limits.

5 0
3 years ago
Read 2 more answers
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