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3 years ago

What the number is 1/10 the value of 237

Mathematics

1 answer:

Troyanec [42]3 years ago

4 0

1/10 the value of 237 means we must multiply 1/10 by 237

1/10 * 237 = 23.7

Hope this helps!

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Please and thankyou (:

den301095 [7]

Let's evaluate the ratio test:

$\dfrac{a_n}{a_{n-1}}=\dfrac{3^{n+1}}{3n+1}\cdot\dfrac{3(n-1)+1}{3^n}=\dfrac{3\cdot3^n}{3n+1}\cdot\dfrac{3n-2}{3^n} = \dfrac{9n-6}{3n+1}$

The limit as $n \to\infty$ of this quantity is 3, which is more than 1. So, the series diverges.

6 0

3 years ago

Hellllllllpppppppppppppppppppppppp

Stolb23 [73]

C is the answer because you can’t square root it any further.

5 0

3 years ago

Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A What is the value of the A^2?

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$

$\boxed{ \tt \red{C}arry \: \red{ O}n \: \red{L}earning} \: \underline{\tt{5/13/22}}$

4 0

2 years ago

75(1+0.2/12)^10*12 what is the answer

Alchen [17]

Answer:

545.119

Step-by-step explanation:

75 * (1 + .2/12)^10 * 12

75 * (1.016667)^10*12

Now things get a little tricky. What does 10*12 mean? I'm taking it to mean 120

75 * (1.016667)^120

75 * 7.268

545.119

8 0

2 years ago

Read 2 more answers

I need help finding x! Please show work for both!

Alenkinab [10]

1. Answer x=-5
You would subtract 25 from each side of the equal sign, taking 25 away from 40 and 25 away from 25, isolating the x. So you would then have -3x=15. You would then divide each side by -3, and getting x=-5

4 0

2 years ago

Read 2 more answers