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3 years ago

−10 > − 5 or −4 < −20

Mathematics

1 answer:

natima [27]3 years ago

3 0

-10 is less than -5 because negative numbers are farther from 0. Same applies to -4 and -20.

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3 (182 + 2) – 172 = -17

IrinaK [193]

Answer:

380≠−17

Step-by-step explanation:

3(182+2)−172 ?= −17

380≠−17

5 0

3 years ago

Read 2 more answers

Twelve pounds less than twice janes weight is 270 pounds. What is Janes weight?

Tpy6a [65]

Janes weight = x
Twelve pounds less than twice janes weight is 270 pounds
12 less than 2x is 270

2x -12 = 270 (add 12 to each side)
2x = 270 + 12
2x = 282 (divide 2 from each side)
x = 282/2
x = 141

The answer is 141 pounds

7 0

2 years ago

PLEASE HELP WITH QUESTIONS 4 AND 5

agasfer [191]

Answer:

4 - Complementary

5 - Supplementary

Step-by-step explanation:

i can't find the value of X

6 0

2 years ago

How do u do this promble

kupik [55]

Answer:

x = 10

Step-by-step explanation:

The total measures of a circle must add up to 360°. In the diagram given there are two angles that are not given, however, both of these should be equal to each other. That means that the sum of the other two angles ('5x - 5' and 93°) must be equal to the other angle of the same measure (138°):

5x - 5 + 93 = 138

Combine like terms: 5x + 88 = 138

Subtract 88 from both sides: 5x + 88 - 88 = 138 - 88 or 5x = 50

Divide by 5: 5x/5 = 50/5 or x = 10

7 0

2 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})$\end{minipage}}$

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

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

brainly.com/question/28155016

3 0

1 year ago