Solve the inequality, round to the nearest tenths

Mathematics

2 answers:

White raven [17]3 years ago

4 0

Answer: x ≥ 10.833333

Step-by-step explanation:

babymother [125]3 years ago

4 0

X ≥ 10.8
Rounded to the nearest tenth

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The length of a rectangle is 5 centimeters more than the width. The area of the rectangle is 36 centimeters. What is the length?

UNO [17]

We can represent the width as x and the length as x+5. Then we can make the area formula A=(x)*(x+5), plug in 36 for A, and solve for x

36=x^2+5x (distribute)
0=x^2+5x-36
0=(x+9)(x-4)
So x=-9 and 4 but dimensions can't be negative so x=4

So if the width is 4, then the length is 9 (4+5)

Hope this helps :)

3 0

3 years ago

Read 2 more answers

Select the correct answer. Solve the following equation by completing the square. 1/4x^2 + x + 1/4 = 0

nekit [7.7K]

Answer:

1/4x^2 + x =-1/4

=−2±√3

8 0

3 years ago

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<img src="https://tex.z-dn.net/?f=cos%20%5B%20arctan%28%5Cfrac%7B12%7D%7B5%7D%29%20-%20arcsin%20%28%5Cfrac%7B-3%7D%7B5%7D%29%5D"

qwelly [4]

Answer: -16/65

Step-by-step explanation:

Drawing the right triangle (as attached) gives us that $\arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)$

Also, $-\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)$

This means our original expression is equal to:

$\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]$

Using the cosine addition formula, which states $\cos(a+b)=\cos a \cos b-\sin a \sin b$ , we get this itself is equal to:

$\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)$

Since $\sin^{2} \theta+\cos^{2} \theta=1$ , we know that:

$\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}$

Similarly, cos(arcsin(3/5))=4/5.

This means the given expression is equal to:

$\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}$