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3 years ago

How many degrees in a half rotation

Mathematics

1 answer:

anygoal [31]3 years ago

5 0

In this half rotation, it is 180 degrees.

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What is the value of x? Enter your answer in the box

USPshnik [31]

It is equilateral triangle, so the 3 sides have the same length

5x - 22 = 4x - 10
5x - 4x = -10 + 22
x = 12

The value of x is 12

8 0

3 years ago

Your grades on three exams are 80, 93, and 91. What grade do you need on the next exam to have an average of 90 on the four exam

Pani-rosa [81]

<u>Answer:</u>

<u>Step-by-step explanation:</u>

Let's assume that the marks needed in the next exam is $x$ . Then if the average becomes 90,

$\frac{80 + 93 + 91 + x}{4}= 90$

Now we can simply solve for $x$ :

⇒ $\frac{264 + x}{4}= 90$

⇒ $264 + x = 360$ [Multiplying both sides by 4]

⇒ $x = 360 - 264$ [Subtracting 264 from both sides]

⇒ $x = \bf96$

Therefore, you need to score 96 in your next exam to have an average of 90 on the four exams.

8 0

2 years ago

9. A boy rides away from home in an automobile at the rate of 28 miles an hour and walks back at the rate of 4 miles an hour. Th

muminat

Answer:

9. A loebskesşbxksjlejs

4 0

2 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

Does this seem answer seem correct?

Blababa [14]

I mean I think so I’m not completely sure it’s a really hard question

5 0

3 years ago