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3 years ago

A stock falls 7 points each day for 2 days. What integer represents the change in stock price for the two days?

1 answer:

5 0

Well the answer would be 14 because the stock fell for 2 days and it falls 7 points each day therefore the answer that is an integer is 14

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PQRS is a rhombus with PQ= 12 cm

Tomtit [17]

Is there an image you can provide if possible

5 0

3 years ago

Use slope and y intercept to graph a line. y=2x-5

Solnce55 [7]

On your graph:

-- Mark a dot on the y-axis, at y = -5 .

-- From there, move 1 unit to the right and 2 units up and make a mark. If this
is too tiny for you, then you can move 7 units to the right and 14 units up, or
11 units to the right and 22 units up ... any way you want to do it, as long as
the distance 'up' is double the distance to the right, because the slope is 2 to 1.
Wherever you wind up, mark a dot.

-- Using your pencil and your ruler, draw a straight line between the two dots you have
marked. You may extend it as far as you wish in either or both directions.

8 0

3 years ago

Read 2 more answers

Kindly help me!!!!!!

Oxana [17]

Answer:

Step-by-step explanation:

AS we can see the lines are parallel so

2 ( 4x - 3) + 7(x + 3) = 180° ( being so - interior angles)

8x - 6° + 7x + 21° = 180°

15x + 15° = 180°

15x = 180° - 15°

15x = 165°

x = 165° / 15

Therefore x = 11°

Now

2 ( 4x - 3) = 2 ( 4 * 11° - 3°) = 2 ( 44 - 3)° = 2* 41 = 82°

7(x + 3 ) = 7 ( 11° + 3°) = 7 * 14 = 98°

4 0

3 years ago

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

Mr Smith bought x number of shirts for the new members of her chorus. The cost for x number of shirts, including $3.99 shipping

fiasKO [112]

We are given that the cost of each shirt = $12.25

It is given that she bought total 'x' shirts, each of cost $12,25.

So, the cost of 'x' shirts is dollars

Also, there is a shipping charge of $3.99

So, the total cost of 'x' shirts is

Since, the shirts cost her total $77.49.

We have the equation,

12.25x + 3.99 = 77.49

4 0

3 years ago