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nadya68 [22]
3 years ago
15

D.) is 13/5 it got cut off HELP NEEDED [10 points up!]

Mathematics
2 answers:
vodomira [7]3 years ago
5 0

It would be A. You can find this out by using SOH CAH TOA,

SOH means sin= opposite/hypotenuse

CAH mean cos= adjacent/hypotenuse

TOA means tangent= opposite/adjacent

You would use CAH since you are trying to find cos∠C

You would have cos∠C=10/26.

26 is the hypotenuse since it is opposite the right angle and is the biggest side.

10 is the adjacent since it is right next to C

You would simplify this down to 5/13 by dividing by 2.

This gives you A.

-Dominant- [34]3 years ago
5 0
<h3>Answer:</h3><h3>A. 5/13</h3>

Hope this helps you!

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1,5,9,13 generalize the pattern by finding the nth term
ladessa [460]

Answer:

an= a1 +(n-1)*4

Step-by-step explanation:

a1 = 1, a2 =5, a3 = 9, a4= 13,... an= ?

we see that each term is getting bigger so usually that will happen because of addition or multiplication-in our case add 4 to previous term

a1 = 1,

a2 = a1+(2-1) 4 = 1+4 = 5,

a3 = a1 + (3-1)*4 = a1+8 = 1 + 8 = 9,

a4= a1 + (4-1)*4 = a1 + 12 = 1+12= 13,

...

an= a1 +(n-1)*4

7 0
3 years ago
Antoine wants to get a subscription to a local library. There are two libraries, each of which charges a monthly subscription fe
Elis [28]

Answer:

2 15.50 50

Step-by-step explanation:

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3 0
3 years ago
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A company’s total cost from manufacturing and selling x units of their product is given by: y = 2x^2 – 600x + 49,000. How many u
AysviL [449]

Answer:

number of units for cost to be minimum=150

Step-by-step explanation:

y=2x^2-600 x+49000

dy/dx=4x-600

dy/dx=0 gives 4x-600=0

4x=600

x=150

d^2y/dx^2=4x

at x=150,d^2y/dx^2=4*150=600>0

so y is minimum at x=150

3 0
4 years ago
Find the vectors T, N, and B at the given point. r(t) = &lt; t^2, 2/3t^3, t &gt;, (1, 2/3 ,1)
maxonik [38]

Answer with Step-by-step explanation:

We are given that

r(t)=< t^2,\frac{2}{3}t^3,t >

We have to find T,N and B at the given point t > (1,2/3,1)

r'(t)=

\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1

T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}

Now, substitute t=1

T(1)=\frac{}{2+1}=\frac{1}{3}

T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}

T'(1)=-\frac{4}{9}+\frac{1}{3}

T'(1)=\frac{1}{9}=

\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}

N(1)=\frac{T'(1)}{\mid T'(1)\mid}

N(1)=\frac{}{\frac{2}{3}}=

N(1)=

B(1)=T(1)\times N(1)

B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}

B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})

B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k

B(1)=\frac{1}{3}

5 0
3 years ago
I need help please help
sveticcg [70]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
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