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3 years ago

Need it please help me

Mathematics

1 answer:

Fynjy0 [20]3 years ago

3 0

The answer to the question is the third one I’m pretty sure

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Can you please answer this question

tamaranim1 [39]

$3tan^{2} \theta +7sec\theta=3$
First I converted the equation terms into sine and cosine.
$tan^{2}\theta = \frac{sin^{2}\theta}{cos^{2}\theta}$ and $sec\theta= \frac{1}{cos\theta}$
Substitution:
$\frac{3sin^2\theta}{cos^2\theta} + \frac{7}{cos\theta} =3$
Common Denominator Created:
$\frac{3sin^2\theta}{cos^2\theta} + \frac{7cos\theta}{cos^2\theta} =3$
Multiply each term by the LCD:
$3sin^2\theta+7cos\theta=3cos^2\theta$
Substitution: Recall ⇒ $sin^2\theta =1-cos^2\theta$
$3(1-cos^2\theta)+7cos\theta=3cos^2\theta$
Distribute and collect all terms on one side:
$6cos^2\theta-7cos\theta-3=0$
Factor and set each factor equal to 0:
$(2cos\theta-3)(3cos\theta+1)=0$
$2cos\theta-3=0$ ⇒ $theta=cos^{-1} \frac{3}{2}$
$3cos\theta+1=0$ ⇒ $theta=cos^{-1} \frac{-1}{3}$
The 2nd factor provides only possible answer 109.5 degrees

4 0

3 years ago

Due tomorrow i will like!

arsen [322]

His weekly allowance is 14$

8 0

3 years ago

Question 2 pleaseee !!!!!!

Leya [2.2K]

It’s 40 degrees
I subtracted 180 minus the 150 and got 30 then addd 110 and 30 and got 140 then substractef that from 180 and got 40

4 0

4 years ago

a customer went to a garden shop and bought some potting soil for $11.50 and 9 shrubs. the total bill was $94.75. write and solv

Mrac [35]

To answer this problem, let p be the price of each scrub. The total bill (t) is the sum of the price of the potting soil (s) and of the 9 scrubs (n). This is depicted in the equation,
$11.50 + 9p = $94.75
The value of p from the equation is 9.25. Thus, each scrub costs $9.25. The answer is letter C.

8 0

3 years ago

Read 2 more answers

Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20

Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

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

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

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

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

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

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago

Need it please help me​

Need it please help me