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3 years ago

I WILL MAKE YOU BRAINLIEST EASY QUESTION

Mathematics

2 answers:

Anni [7]3 years ago

5 0

Answer:

Hello,

; 1.38; 1.4; 1 3/7

Step-by-step explanation:

skelet666 [1.2K]3 years ago

3 0

Answer:

D.)

Step-by-step explanation:

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Six times a number, increased by 3 is is 27. find the number

Mekhanik [1.2K]

Answer : The number is 4.

Step-by-step explanation:

Given : 

six times a number, increased by 3 is is 27.

To Find: 

The number.

Solution : 

Let the number is x

six times a number is 6x

Increased by 3 = 6x+3

According to the given question :

Solving the value x : 

6x+3 =27

6x = 27-3

6x = 24

x = 24/6

x = 4

Hence, the number is 4.

6 0

3 years ago

A picture is 6 inches longer than it is wide and has an area of 167

Scorpion4ik [409]

Answer:

A) L = W + 6

B) L * W = 167 OR L = 167 / W

Substituting B) into A)

167 / W = W + 6

Multiplying both sides by "W" we get

167 = W^2 + 6W

W^2 + 6W -167

Solving by the quadratic formula

Width = 10.266

Length = 16.266

Step-by-step explanation:

5 0

3 years ago

Right angle are known as the

marissa [1.9K]

Right angle are known as exactly 90 degrees corresponding to a quarter turn.

8 0

2 years ago

Read 2 more answers

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

Which of the following could be a measurement on a scale drawing that has a scale of 2:5?

klemol [59]

Answer:

Step-by-step explanation:

3 0

2 years ago