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Furkat [3]
2 years ago
10

6. Determine which of the following binomials is a factor of P(x) = x - 7x + 5. c. x+2 a. X-1 b. x + 1 d. none of these 7. Which

is the missing factor in the equation x² – 4 = (x - 2) .)? a.x-2 c. x + 4 b. x + 2 d. x-4 PREPARED BY:​
Mathematics
1 answer:
malfutka [58]2 years ago
4 0

I'm Sorry I can't answer the question because it's messed up and I haven't learned that yet

Maybe you can try another app

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The difference of two numbers is 16 and the sum of their squares is 130. What are the numbers.
Korvikt [17]
x - y = 16
x^2 + x^2 = 130

----------------
x^2 - y^2 = 256
+(x^2 + y^2 = 130)
2x^2 = 386
x^2 = 193
x = <span>13.892 (3DP)
----
</span>x - y = 16
13.892 - y = 16
-y = 2.108
y = -2.108
The numbers are: 13.892 and -2.108
4 0
3 years ago
If 6/9 is equal to 10/R, what is R?
sladkih [1.3K]

Answer:

r = 15

Step-by-step explanation:

I don't really have an explanation, I'm just good at quick math,

I'm sorry if you wanted an explanation, I would explain but I am really bad at explaining math.

5 0
2 years ago
The distance by road from town A to town B is 257 km. What is 50% of that distance?
Marina86 [1]

Answer: 128.5km

Step-by-step explanation:

Since we are given the information that the distance by road from town A to town B is 257 km. To get 50% of the distance, we simply have to multiply the distance given by 50%. This will be:

= 50% × 257km

= 50/100 × 257km

= 0.5 × 257km

= 128.5km

Therefore, 50% of the distance is 128.5km.

3 0
2 years ago
A bag contains pennies nickels dimes and quarters. There are 50 coins in all. Of the coins 14 percent are pennies and 42 percent
lord [1]
I believe it’s $5.47
5 0
3 years ago
Find a power series representation of e^x sin(x)
liberstina [14]
The Taylor series is defined by:
f(x) = \sum \frac{f^n (a)}{n!} (x-a)^n

Let a = 0.
Then its just a matter of finding derivatives and determining how many terms is needed for the series.

Derivatives can be found using product rule:
f^n (x) = e^x g^{n-1} (x) + e^x g^n (x)  \\ g^0 (x) = sin x

Do this successively to n = 6.
f^1 (x) = e^x (sinx +cos x) \\ f^2 (x) = e^x (2cos x)  \\ f^3 (x) = e^x(2cos x - 2sin x)  \\ f^4 (x) = e^x (-4 sin x)  \\ f^5 (x) = e^x(-4sinx -4cos x) \\ f^6(x) = e^x(-8cos x)

Plug in x=0 and sub into taylor series:
e^x  sin x = x+x^2 +\frac{x^3}{3} -\frac{x^5}{30}-\frac{x^6}{90}...

If more terms are needed simply continue the recursive derivative formula and add to taylor series.
6 0
3 years ago
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