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WARRIOR [948]
2 years ago
8

47 + (-13 + 7)- (-13 -6)

Mathematics
1 answer:
Andrei [34K]2 years ago
6 0

Answer:

60

Step-by-step explanation:

(-13 + 7) = -6

(-13 -6) = -19

-6 - -19 =13

47+13 =60

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Please can someone explain how to work out this
wariber [46]

Answer:

T = 4B + 10C

Step-by-step explanation:

Since there are 4 doughnuts in a bag then there are 4B doughnuts in B bags

Since there are 10 doughnuts in a carton then there are 10C doughnuts in C cartons, thus summing the two quantities gives

T = 4B + 10C

6 0
3 years ago
In a simple regression analysis (where y is a dependent and x an independent variable), if the slope is positive, then it must b
Scorpion4ik [409]

Answer:

c. there is a positive correlation in-between x and y

Step-by-step explanation:

A regression line is a line that suggests that all the points in a scatter diagram lie on or near one particular line. In a simple regression analysis in which y is the dependent variable and x is the independent variable. If the slope is positive, the bivariate data is also said to have a positive correlation. The positive correlation in-between two variables x and y implies that in general, an increase in x goes hand in hand with an increase in y.

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2 years ago
0.13x0.11=????????????????
andre [41]

Answer:

0.0143 in decimal form

Step-by-step explanation:

3 0
3 years ago
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Natalka [10]
Answers B and C are true.
6 0
3 years ago
Read 2 more answers
Help me with differentation and integration please!!
Marina86 [1]

Answer:

See below

Step-by-step explanation:

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

Recall

\dfrac{d}{dx}\tan x=\sec^2

Using the chain rule

\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}

such that u = \tan x

we can get a general formulation for

y = \tan^n x

Considering the power rule

\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}

we have

\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x

therefore,

\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x

Now, once

\sec^2 x - 1= \tan^2x

we have

3\tan^2x \sec^2x =  3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x

Hence, we showed

\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x

================================================

For the integration,

$\int \sec^4 x\, dx $

considering the previous part, we will use the identity

\boxed{\sec^2 x - 1= \tan^2x}

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx $

Considering u = \tan x

and then du=\sec^2x\ dx

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx  = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0
2 years ago
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