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Ronch [10]
3 years ago
8

What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?

Mathematics
1 answer:
RideAnS [48]3 years ago
6 0

Step-by-step explanation:

The function is as follows :

f(x) = x^2 + 5x + 5

We need to find the zeroes of the function in the simplest radical form.The zero of the above function is given by :

x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here,

b = 5

a = 1

c = 5

So,

x=\dfrac{-5\pm \sqrt{5^2-4\times 1\times 5} }{2(1)}\\\\x=\dfrac{-5\pm \sqrt{25-20} }{2(1)}\\\\x=\dfrac{-5\pm \sqrt{5} }{2}

Hence, the correct option is (c).

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Help please, thanks.
Vladimir79 [104]

Answer:

B

Step-by-step explanation:

C = R/1.8 - 273

C + 273 = R/1.8

1.8 * (C + 273) = R

1.8 can also be read as the fraction 9/5.

9/5 * (C + 273) = R

Thus your answer is B. Hope I helped!

7 0
3 years ago
Read 2 more answers
What number is 9 times as much as 400 ? A 391 B 409 C 3,600 D 3,609​
Eddi Din [679]

Answer:

C.3600

...............

8 0
2 years ago
19. Use Gauss-Jordan elimination to solve the following system of equations. 3x + 5y = 7 6x − y = −8
MrRa [10]

\left[\begin{array}{cc}3&5\\6&-1\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}7\\-8\end{array}\right]\\\left[\begin{array}{cc|c}3&5 &7\\ 6& -1 & -8\end{array}\right] \rightarrow \left[\begin{array}{cc|c}3&5& 7\\0&-11 & -22\end{array}\right] \rightarrow \left[\begin{array}{cc|c}3&5& 7\\0&1 & 2\end{array}\right]\\3x +5y = 7\\\\y = 2\\\therefore x = -1

6 0
3 years ago
elijah eats at his favorite restaurant 4 times a week. He buys a sandwich for $8 and a drink for $2.Write an equation that can b
mario62 [17]
If he goes there 4 times a week then $8 plus $2 equals $10 and times $10 by 4 times a week equals $40

So Elijah spends $40 a week at the restaurant
5 0
3 years ago
Read 2 more answers
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
3 years ago
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