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denis-greek [22]
3 years ago
10

Find the value of x (Round to the nearest tenth) Help plssssssssss

Mathematics
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

bbhwhehwyayyquqysyegegegsgsgsgsusus

Step-by-step explanation:

vgsgahah z z zb

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900 is 4 1/2% of what?
Svet_ta [14]
Simple, first change the 4.5% into 0.045

Thus,

900=0.045*n

900/0.045=0.045n/0.045

20,000=n

Thus, your answer is 20,000.
3 0
3 years ago
Read 2 more answers
Hi, teacher I was absent these days and I didn’t understand anything about this lesson and I need help this is not count as a te
motikmotik

Given:

There are given that the cos function:

cos210^{\circ}=-\frac{\sqrt{3}}{2}

Explanation:

To find the value, first, we need to use the half-angle formula:

So,

From the half-angle formula:

cos(\frac{\theta}{2})=\pm\sqrt{\frac{1+cos\theta}{2}}

Then,

Since 105 degrees is the 2nd quadrant so cosine is negative

Then,

By the formula:

\begin{gathered} cos(105^{\circ})=cos(\frac{210^{\circ}}{2}) \\ =-\sqrt{\frac{1+cos(210)}{2}} \end{gathered}

Then,

Put the value of cos210 degrees into the above function:

So,

\begin{gathered} cos(105^{\circ})=-\sqrt{\frac{1+cos(210)}{2}} \\ cos(105^{\operatorname{\circ}})=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ cos(105^{\circ})=-\sqrt{\frac{2-\sqrt{3}}{4}} \\ cos(105^{\circ})=-\frac{\sqrt{2-\sqrt{3}}}{2} \end{gathered}

Final answer:

Hence, the value of the cos(105) is shown below:

cos(105^{\operatorname{\circ}})=-\frac{\sqrt{2-\sqrt{3}}}{2}

4 0
1 year ago
High of the following transformations does not result in a congruent figure
LuckyWell [14K]
Any transformation involving change of scale will not result in a congruent figure. Rotations and reflections and translations maintain congruence.
7 0
4 years ago
Consider this equation. cos(theta) = (-2√5)/5. If theta is an angle in quadrant II, what is the value of sin(theta)? NO LINKS!!!
vlada-n [284]
  • In second quadrant sin and cosec are positive

\\ \tt\hookrightarrow sin^2\theta=1-cos^2\theta

\\ \tt\hookrightarrow sin^2\theta=1-(\dfrac{-2\sqrt{5}}{5})^2

\\ \tt\hookrightarrow sin^2\theta=1-\dfrac{20}{25}

\\ \tt\hookrightarrow sin^2\theta=\dfrac{5}{25}

\\ \tt\hookrightarrow sin\theta=\dfrac{\sqrt{5}}{5}

\\ \tt\hookrightarrow sin\theta=\dfrac{1}{\sqrt{5}}

5 0
3 years ago
Find a general solution of y" + 8y' + 16y=0.
Misha Larkins [42]

Answer:

The general solution: C_{1}e^{-4x} + xC_{2}e^{-4x}

Step-by-step explanation:

Differential equation: y'' + 8y' + 16y = 0

We have to find the general solution of the above differential equation.

The auxiliary equation for the above equation can be writtwn as:

m² + 8m +16 = 0

We solve the above equation for m.

(m+4)² = 0

m_{1} = -4, m_{2} = -4

Thus we have repeated roots for the auxiliary equation.

Thus, the general solution will be given by:

y = C_{1}e^{m_{1}x} + xC_{2}e^{m_{2}x}

y = C_{1}e^{-4x} + xC_{2}e^{-4x}

6 0
3 years ago
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