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2 years ago

Here this is a giveaway free 100 i started back

Mathematics

2 answers:

Elenna [48]2 years ago

4 0

Answer:

Step-by-step explanation:

Yakvenalex [24]2 years ago

3 0

Answer:

thanks!

Step-by-step explanation:

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Mrs Belen wanted to bake 10 loaves of bread but realized she needed 1.75 more of flour. If she baked 7 loaves of bread she would

Pani-rosa [81]

To answer this question, follow this step by step procedure:1. Add first the needed and the excess flour to determine to total flour, = (1.75 kg + 0.5 kg = 2.25)2. Then divide the answer by 3 since you need to have 3 more breads, 2.25 kg / 3 = 0.75 kg
So we can say that we need 0.75 kilogram of flour for each loaf of bread.

3 0

3 years ago

Look at photo (WITH THE PHOTO)

frez [133]

The answer to that would be 3/8

8 0

2 years ago

Read 2 more answers

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

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

How do i solve this pls

miskamm [114]

First we would solve both x’s
3x^2 - 5x
But 3 has the thingy thing (forgot the name haha)
Which means you have to multiply it by itself that many times which would be
3x3=9
So now it would be
9x- 5x = 4x
Which leaves 7 and 4x

Now you may subtract those two

7-4x= 3x

7 0

2 years ago

Dylan wants to buy both the tablet and

nignag [31]

Answer:

108$

Step-by-step explanation:

6 0

2 years ago