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3 years ago

Here is an inequality: 7x+6/2 < 3x +2

Mathematics

1 answer:

Archy [21]3 years ago

7 0

Answer:

x=1

Step-by-step explanation:

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If 't' bags of chips cost $c, what is the cost of 5 bags of chips?

allsm [11]

Answer:

$ (C/t ) × 5

= $ 5C/t

.....................................

4 0

2 years ago

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

$-\dfrac{7}{25}$

Step-by-step explanation:

<u>Trigonometric Identities</u>

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

<u>Trigonometric ratios</u>

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos(\angle ABC)=\dfrac{3}{5}$

$\sin(\angle ABC)=\dfrac{4}{5}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}$

7 0

2 years ago

Read 2 more answers

How do you find perimeter

Veronika [31]

All you do is add up all your sides! So if your shape has 4+4+5+5 = 18. All you do is add them up!

3 0

3 years ago

Evaluate the integral following

alina1380 [7]

Answer:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

Step-by-step explanation:

We are given the integral of:

$\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}$

First, we can use a property to separate a constant out of integrand:

$\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}$

Next, expand the expression (integrand):

$\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}$

Since $\displaystyle{\sec x = \dfrac{1}{\cos x}}$ then it can be simplified to:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}$

Recall the formula:

$\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}$

For $\displaystyle{\cos ^2 x}$ , we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

$\displaystyle{2\cos^2 x -1 = \cos2x}$

We can solve for $\displaystyle{\cos ^2x}$ which is:

$\displaystyle{2\cos^2 x = \cos2x+1}\\\\\displaystyle{\cos^2x = \dfrac{\cos 2x +1}{2}}$

Therefore, we can write new integral as:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \dfrac{\cos2x +1}{2} \, dx}$

Evaluate each integral, applying the integration formula:

$\displaystyle{\int \dfrac{1}{\cos^2x} \, dx = \boxed{\tan x + C}}\\\\\displaystyle{\int -2 \, dx = \boxed{-2x + C}}\\\\\displaystyle{\int \dfrac{\cos 2x +1}{2} \, dx = \dfrac{1}{2}\int \cos 2x +1 \, dx}\\\\\displaystyle{= \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2x + x\right) + C}\\\\\displaystyle{= \boxed{\dfrac{1}{4}\sin 2x + \dfrac{1}{2}x + C}}$

Then add all these boxed integrated together then we'll get:

$\displaystyle{4\left(\tan x - 2x + \dfrac{1}{4}\sin 2x + \dfrac{1}{2} x\right) + C}$

Expand 4 in the expression:

$\displaystyle{4\tan x - 8x +\sin 2x + 2 x + C}\\\\\displaystyle{4\tan x + \sin 2x - 6x + C}$

Therefore, the answer is:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

4 0

1 year ago

What is the area of a rectangle with vertices at (−4, 0) , (−3, 1) , (0, −2) , and (−1, −3) ?

skad [1K]

Answer: $6\ units^2$

Step-by-step explanation:

Plot the vertices of the rectangle on a coordinate plane (Observe the figure attached where the rectangle is identified as ABCD).

The area of a rectangle can be calculated with this formula:

$A=lw$

Where "l" is the lenght and "w" is the width.

In order to find the lenght and the width, you can use the formula for calculate the distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Knowing the coordinates of the vertices, you get:

$w=d_{AB}=\sqrt{(-4-(-3))^2+(0-1)^2}=\sqrt{2}\ units\\\\l=d_{BC}=\sqrt{(0-(-3))^2+(-2-1)^2}=3\sqrt{2}\ units$

Therefore, substituting values into the formula $A=lw$ , you get that the area of the rectangle is:

$A=(3\sqrt{2}\ units)(\sqrt{2}\ units)=6\ units^2$

8 0

3 years ago