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Allisa [31]
3 years ago
6

BRAINLIEST ASAP! PLEASE HELP ME :)

Mathematics
1 answer:
masha68 [24]3 years ago
5 0

Answer:

○\displaystyle 0 + 2kπ

Explanation:

<em>See above</em><em> </em><em>explanation</em>

I am joyous to assist you anytime.

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What is the sum of the first 11 terms of the arithmetic series in which a1 = 12 and d = 5?. (Points : 1). 38. . 549. . 143. . 2
Marina CMI [18]
First we will find the 11th term
an = a1 + (n-1) * d
a11 = 12 + (11 - 1) * 5
a11 = 12 + 10 * 5
a11 = 12 + 50
a11 = 62

now we use the sum formula...
Sn = (n (a1 + an)) / 2
S11 = (11 (12 + 62)) / 2
S11 = (11 (74)) / 2
S11 = 814/2
S11 = 407
3 0
2 years ago
Read 2 more answers
I really don't understand why this question was deleted its not bad I don't know how to explain it
nalin [4]

Answer:

Step-by-step explanation:

Because it is a square root

8 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
3 years ago
How do I solve these equations
Sever21 [200]
You need to know the properties of each function.
Tan is opposite (y) over adjacent (x).  Sin>0 means that sin is positive, therefore, it is located on either quadrant 1 or 2. Tan=4/3  so it is located in quadrant 1.
The side of the triangle must be Adjacent=3  opposite=4 and hypotenuse=5
Now, you are asked to find the half angle of Cos which is 
\sqrt{\frac{1+cos}{2} }
By following the formula, cos=3/5 then:
\sqrt{ \frac{1- \frac{3}{5} }{2} }    Multiply everything (inside the square          \sqrt{ \frac{5+3}{10} }                  root) by 5
\frac{2 \sqrt{2} }{ \sqrt{10} }
\frac{2 \sqrt{20} }{10}                Then just simplify
\frac{ \sqrt{20} }{5}      The answer is Square root of 20 over 5
3 0
3 years ago
22<br> 1 point<br> If f (x) = 7– 41, what is f (-10)?
Anettt [7]

Answer:

f(x)+41 f(-10)=7

Step-by-step explanation:

5 0
3 years ago
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