All categories

3 years ago

What is the answer?

Mathematics

1 answer:

Mrrafil [7]3 years ago

6 0

Answer:

53°

Step-by-step explanation:

The angle that measures 143 is vertical to the right angle and the ? angle.

Vertical angles have the same measure. <? + <90 = <143

So do 143 - 90 = 53

So the ? angle = 53

Hope this helps ya!!

You might be interested in

A worm travels 1,250 millimeters.How many meters did the worm travel

vazorg [7]

The answer is 1.25 meters.

hope it helps

can you choose mine as the brainliest answer

5 0

3 years ago

Make t the subject of the formula p = 4t+1/2-t

alex41 [277]

Answer:

t = (p - 1/2)/3

Step-by-step explanation:

p = 4t + 1/2 - t

combine like terms:

p = 3t + 1/2

subtract 1/2 from each side of the equation:

3t = p - 1/2

divide both sides by 3:

t = (p - 1/2)/3

8 0

2 years ago

What is 34 & 3/4 inches divided by 2 equal

Sever21 [200]

Answer:

17 and 3/8 inches

Step-by-step explanation:

..............................

(34+3/8)/2=17+3/16

plz....

mark it as a brilliant answer

7 0

2 years ago

Jason complete 5/8 of his test and 10/25 of his homework. What fraction of his homework does he still need to complete?

sineoko [7]

Answer:

3/4

Step-by-step explanation:

Because the total homework would be 25/25

$\frac{25}{25}-\frac{10}{25} =\frac{15}{25}$

15/20 can be reduced to 3/4

4 0

3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$