Help meplease help me these problems

Mathematics

1 answer:

natima [27]3 years ago

4 0

15) 6.1

17) 0.91

19) 86.96

20) I'm not sure,but I think is 82.1%

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Find the measure of angle C of a triangle ABC, if angle A=a and angle B= 2a.

erma4kov [3.2K]

3.14 - 3a Radians

Because
180o = 3.14 radian

3 0

2 years ago

What os the answer to 14÷2.5=

almond37 [142]

Answer:

5.6

Step-by-step explanation:

by dividing

6 0

2 years ago

Read 2 more answers

Solve the equations if possible * -5×=×+6(1-×) *(2×-5)²=(2×-1)(2×+1)-10(2×-1)

Allushta [10]

-5x = x + 6(1-x)

-5x = x + 6(1) + 6(-x)
-5x = x + 6 -6x
-5x = x - 6x + 6
-5x = -5x + 6
-5x + 5x = 6
0 = 6 Not equal. No solution.

(2x-5)² = (2x-1)(2x+1) -10(2x-1)
(2x-5)(2x-5) = (2x-1)(2x+1) - 20x + 10
2x(2x-5)-5(2x-5) = 2x(2x+1)-1(2x+1) - 20x + 10
4x² - 10x -10x + 25 = 4x² + 2x - 2x -1 - 20x + 10
4x² - 20x + 25 = 4x² - 20x - 1 + 10
4x² - 20x + 25 = 4x² - 20x - 9

Not equal. No solution.

8 0

2 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$