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2 years ago

Let m=4 and n=36. What is the value of m+.9?

1 answer:

3 0

Answer:
4.9

Explanation:
We are given m = 4 and n = 36 and the expression m + .9.

Replace m with the given value:
4 + .9

Add them together:
4.9

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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})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

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

Learn more about indefinite integration here:

brainly.com/question/27805589

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3 0

2 years ago

Please answer hurry!!!!

lutik1710 [3]

The best way to solve this problem is by plugging in each answer choice for both equations and seeing which one works. You can already see that choice A doesn't work because -1 is not greater than or equal to 4. That means you just need to test choices B and C!

Choice B (4, -2)
$x + y \geq 6\\
4 + -2 \geq 6\\
2 \geq 6 
$
2 is not greater than or equal to 6, so choice B doesn't work.

Choice C (4, 2)
$x + y \geq 6\\ 
4 + 2 \geq 6\\ 
6 \geq 6 \\\\
x \geq 4\\
4 \geq 4$
Since 6 is equal to 6 and 4 is equal to 4, choice C works!

-------

Answer: C) (4, 2)

6 0

3 years ago

What is the perimeter of this red

Vilka [71]

96

multiply the side length by the number of sides

4 0

3 years ago

Find midpoint of line segment joining j(6 17 and i(9 16)

Serga [27]

Answer: (15/2, 33/2) or (7.5, 16.5)
depending on if you want fractions or decimals

explanation:
equation for midpoint... (avg x, avg y)
x-coordinate: (6+9)/2 = 15/2 = 7.5
y-coordinate: (17+16)/2 = 33/2 = 16.5

4 0

3 years ago

Multiply the binomials: (w-9) • (w-4)

Novay_Z [31]

Answer:

w² - 13w + 36

Step-by-step explanation:

Given

(w - 9)(w - 4)

Each term in the second factor is multiplied by each term in the first factor, that is

w(w - 4) - 9(w - 4) ← distribute both parenthesis

= w² - 4w - 9w + 36 ← collect like terms

= w² - 13w + 36

3 0

3 years ago

Let m=4 and n=36. What is the value of m+.9?​

Let m=4 and n=36. What is the value of m+.9?