I have a question about fractions

Mathematics

2 answers:

Dima020 [189]2 years ago

8 0

Answer: 17/20 is equivalent to 34/40 or 1 7/10.

yarga [219]2 years ago

5 0

Answer:

please just add the question

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F(x)=( x^4 + 5x^3 - 2x^2 + 5x - 3) d(x)=( x^2 + 1) Divide f(x) by d(x)

VLD [36.1K]

f '(x) = [(-40x +11)(7x - 9) - 7(-4x +3)(5x + 1)]/(7x - 9)2
= [(-280x2 + 360x + 77x - 99) - 7(-20x2 - 4x + 15x + 3)]/(7x - 9)2
= [(-280x2 + 437x - 99) + (140x2 + 28x - 105x - 21)]/(7x - 9)2
= (-140x2 +360x - 120)/(7x - 9)2

i think thats how you would solve it
hope this helps tho:)

4 0

3 years ago

The temperature in one northern city was -12°F and in a southern city was 57°F. How much warmer

lara [203]

Answer

69°F

Because

12+57=69

3 0

3 years ago

F your car gets 2424 miles per gallon, how much does it cost to drive 370370 miles when gasoline costs $3.203.20 per gallon?

TiliK225 [7]

$351,912.62 would be the cost

4 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]

$\large \mathbb{PROBLEM:}$

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

$\large \mathbb{SOLUTION:}$

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