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Arte-miy333 [17]
3 years ago
12

HELP ME PLEASE !!!!!!

Mathematics
1 answer:
Lunna [17]3 years ago
6 0

Answer:

A

Step-by-step explanation:

The answer is (a) because that value is the y intercept meaning that it will be the value for price at the beginning of the time frame... in this case in January 2013 which is the beginning of the year.

Hope you find this helpful.

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Please can anyone help what is AC
Nesterboy [21]

The answer is C.

You need you use Pythagoras theorem 2 times to have 2 equations with the same sides i.e. for example a2 + b2 = 81 and a2 - b2 = 9.

You can do it as you have different triangles but with the same sides.

7 0
3 years ago
1/2 times what =5<br><br><br> I need help pls its due tomorrow. :(
Mashcka [7]

Answer:

10

Step-by-step explanation:

half of 10 is 5

8 0
2 years ago
Read 2 more answers
In ΔTUV, v = 990 inches, t = 840 inches and ∠U=148°. Find the length of u, to the nearest inch.
Scilla [17]

Answer:

the answer is 1760

Step-by-step explanation:

5 0
3 years ago
Find the complex zeros of <br><img src="https://tex.z-dn.net/?f=%20f%28x%29%20%3D%20%7Bx%7D%5E%7B3%20%7D%20%20-%2013%20%7Bx%7D%5
timama [110]

f(x) = 0 \\\\\implies x^3-13x^2+59x -87 =0\\\\\implies x^3 -3x^2 -10x^2 +30x +29x - 87=0\\\\\implies x^2(x-3)  - 10x(x-3) + 29(x-3) =0\\\\\implies (x-3)(x^2 -10x +29) =0\\\\\implies x - 3 = 0 ~~ \text{or}~~x^2 -10x +29 = 0\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{-(-10) \pm \sqrt{(-10)^2 -4 \cdot 1 \cdot 29}}{2(1)}\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{10\pm \sqrt{-16}}{2}\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{10\pm 4i }{2}\\

\\\implies x =3~~ \text{ or} ~~ x =5 \pm 2i\\\\\text{Hence the complex roots are,}~ 5- 2i ~~ \text{and} ~~5 + 2i

6 0
3 years ago
in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference​
allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}

so that

a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d

a_n=a_1+(n-1)d

33=8+(n-1)d

21=(n-1)d

n has to be an integer, which means there are 4 possible cases.

Case 1: n-1=1 and d=21. But

\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123

Case 2: n-1=21 and d=1. But

\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123

Case 3: n-1=3 and d=7. But

\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123

Case 4: n-1=7 and d=3. But

\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123

8 0
3 years ago
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