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3 years ago

Can you solve question #36

Mathematics

2 answers:

blagie [28]3 years ago

6 0

Your answer would be 75 for both base angles, all angles in a triangle will equal 180 when added together. Therefor 75+75+30=180

Sonja [21]3 years ago

5 0

If the measure of the vertex angle is 30 degrees, then the measures of the base angles are 75 degrees.

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Trig proofs with Pythagorean Identities.

lorasvet [3.4K]

To prove:

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Solution:

$$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$

Multiply first term by $\frac{1+cos x}{1+cos x}$ and second term by $\frac{1-cos x}{1-cos x}$ .

$$= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}$

Using the identity: $(a-b)(a+b)=(a^2-b^2)$

$$= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}$

Denominators are same, you can subtract the fractions.

$$= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}$

Using the identity: $1-\cos ^{2}(x)=\sin ^{2}(x)$

$$= \frac{1+\cos^2 x}{\sin^2x}$

Using the identity: $1=\cos ^{2}(x)+\sin ^{2}(x)$

$$=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}$

$$=\frac{\sin ^{2}x+2 \cos ^{2}x}{\sin ^{2}x}$ ------------ (1)

$RHS=2 \cot ^{2} x+1$

Using the identity: $\cot (x)=\frac{\cos (x)}{\sin (x)}$

$$=1+2\left(\frac{\cos x}{\sin x}\right)^{2}$

$$=1+2\frac{\cos^{2} x}{\sin^{2} x}$

$$=\frac{\sin^2 x + 2\cos^{2} x}{\sin^2 x}$ ------------ (2)

Equation (1) = Equation (2)

LHS = RHS

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Hence proved.

5 0

3 years ago

Which ordered pairs are solutions to the inequality

Kipish [7]

Answer:

(1,-1)

(7,12)

(5,-3)

Step-by-step explanation:

we know that

If a ordered pair is a solution of the inequality, then the ordered pair must satisfy the inequality

we have

$y-2x \leq -3$

Verify each case

case 1) we have

(1,-1)

substitute the value of x and the value of y in the inequality and then compare the results

$-1-2(1) \leq -3$

$-3 \leq -3$ ----> is true

therefore

The ordered pair is a solution of the inequality

case 2) we have

(7,12)

substitute the value of x and the value of y in the inequality and then compare the results

$12-2(7) \leq -3$

$-12 \leq -3$ ----> is true

therefore

The ordered pair is a solution of the inequality

case 3) we have

(-6,-3)

substitute the value of x and the value of y in the inequality and then compare the results

$-3-2(-6) \leq -3$

$9 \leq -3$ ----> is not true

therefore

The ordered pair is not a solution of the inequality

case 4) we have

(0,-2)

substitute the value of x and the value of y in the inequality and then compare the results

$-2-2(0) \leq -3$

$-2 \leq -3$ ----> is not true

therefore

The ordered pair is not a solution of the inequality

case 5) we have

(5,-3)

substitute the value of x and the value of y in the inequality and then compare the results

$-3-2(5) \leq -3$

$-13 \leq -3$ ----> is true

therefore

The ordered pair is a solution of the inequality

6 0

3 years ago

Can somebody please show me how to find A? thanks

NNADVOKAT [17]

Hello,
Answer C

Using Thales, 3/8=5/A==>A=5*8/3=40/3

6 0

3 years ago

3(4x+8) + 3(2x - 6)<br> Match the equivalent expressions

slamgirl [31]

Answer:=18x+6

Step-by-step explanation:

3(4x+8)+3(2x-6)

12x+24+6x-18

12x+6x+24-18

18x+6

5 0

3 years ago

Find the ordered pairs for the x- and y-intercepts of the equation 2x − 6y = 24 and select the appropriate option below. (5 poin

Blizzard [7]

Answer: C

Step-by-step explanation:

The x-intercept is found when y=0, and the y-intercept is found when x=0. If y=0, then we have:

2x=24

x=12

Therefore, the x-intercept is 12, so the full coordinates are (12, 0). Note that the only option with this x-intercept is C, so C is the correct answer.

You can find the y-intercept similarly. If x is zero, then:

-6y=24

y=-4

Therefore, (0, -4) is the y-intercept.

7 0

3 years ago