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Maslowich
3 years ago
12

Round 307 to neatest tens

Mathematics
2 answers:
hammer [34]3 years ago
5 0
310 is the nearest tenth
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omeli [17]3 years ago
5 0
The awnser to your question is 310
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What is a decimal as a eighth note
Vera_Pavlovna [14]

Idk,
Tips with decimals....

Our decimal system gives us the flexibility to write numbers as large or small as we like. The key to the decimal system is the decimal point. Anything on the left of the decimal point represents a whole number, anything on the right of the decimal represents less than one (similar to a fraction). Going from left to right, the value of each place on the right of the decimal point is 1/10 the value of the place on the left. Good luck!


4 0
3 years ago
Anyone please HELP cause I’m getting frustrated
pashok25 [27]

Answer:

U =35.5

Step-by-step explanation:

Since this is a right angle, we can use trig functions

tan theta = opp/ adj

tan U = 5/7

Take the inverse of each side

tan ^-1 tan U = tan ^-1 (5/7)

U = 35.53767779

Round to the nearest tenth

U =35.5

4 0
3 years ago
M is directly proportional to the square of X
Oliga [24]

Answer:

m=kx squared

24=k*2squared

24/4=k

k=6

24= 6*4

m=6xsquared

5 0
3 years ago
Suppose r(t)=costi+sintj+3tk represents the position of a particle on a helix, where z is the height of the particle above the g
Ilia_Sergeevich [38]

a. The \vec k component tells you the particle's height:

3t=16\implies t=\dfrac{16}3

b. The particle's velocity is obtained by differentiating its position function:

\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k

so that its velocity at time t=\frac{16}3 is

\vec v\left(\dfrac{16}3\right)=-\sin\dfrac{16}3\,\vec\imath+\cos\dfrac{16}3\,\vec\jmath+3\,\vec k

c. The tangent to \vec r(t) at t=\frac{16}3 is

\vec T(t)=\vec r\left(\dfrac{16}3\right)+\vec v\left(\dfrac{16}3\right)t

4 0
2 years ago
What is the length of EF in the right triangle below?
Vinil7 [7]
Your question does not give enough information for me to answer
8 0
2 years ago
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