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PLEASE HELP ME SOLVE!

Answers

Answer 1

Answer:

Answer:

F(-6) = 2*(-6) + 1

5*(-6) -2

= -11

-32

= 11

Answer 2

Answer: F=-6

step by step explanation :

2(-6)+1
———-
5(-6) -2

-12 + 1
———-
-30 -2

-11
—-
-32

Answer : 11
—
32

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Solve the following system of equations using the elimination method.7x + 20y = 14
2x – 10y = 4
Question 18 options:

A)

(3,1)

B)

(2,0)

C)

(4,–5)

D)

(–3,4)

Answers

Answer:

opt B (2,0)

Step-by-step explanation:

pls rate

Answer:

Step-by-step explanation:

Based on information from the periodic table, what does this image represent? HE 4 Protons e 5 Neutrons = 2 Electrons A. A negatively charged beryllium ion B. A neutral boron atom
C. A neutral beryllium atom
D. A positively charged beryllium ion

Answers

This image represents A positively charged beryllium ion.

We have given that,

He 4 Protons e 5 Neutrons = 2 Electrons

A. A negatively charged beryllium ion

B. A neutral boron atom

C. A neutral beryllium atom

D. A positively charged beryllium ion.

We have to determine, Based on information from the periodic table what does this image represent?

What is the periodic table?

A table of the chemical elements arranged in order of atomic number, usually in rows, so that elements with similar atomic structure (and hence similar chemical properties) appear in vertical columns.

This image represents A positively charged beryllium ion.

To learn more about the periodic table visit:

brainly.com/question/15987580

#SPJ2

Answer:

D. A positively charged beryllium ion

How many fixed points are there for the function f(x)=x^4

Answers

The fixed points in the function f(x) = x^4 are:

x = 0, x = 1, x = -0.5 - 0.866i, and x = -0.5 +0.866i

The given function is:

$f(x) = x^4$

The fixed points of a function f(x) is determined by solving f(x) - x = 0

$x^4 - x = 0\nx(x^3-1) = 0\nx(x-1)(x^2+x+1) = 0$

Equate each of the terms to zero:

x = 0

x - 1 = 0

x = 1

$x^2+x+1 = 0\n\nx = (-b\pm √(b^2-4ac) )/(2a) \n\nx = (-1\pm √(1^2-4(1)(1)) )/(2(1)) \n\nx = (-1\pm √(-3) )/(2)\n\nx= (-1\pm i√(3) )/(2)\n\nx = (-1 \pm i√(3) )/(2)\n\nx= -0.5\pm0.866i\n\nx_1 = -0.5-0.8661\n\nx_2 = -0.5+0.866i$

Therefore, the fixed points in the function f(x) = x^4 are:

x = 0, x = 1, x = -0.5 - 0.866i, and x = -0.5 +0.8661

Learn more here:brainly.com/question/14785053

Answer:

What is this?

Step-by-step explanation:

Kjsdfjyrbasdhviurhasiudhvr = ?

Answers

Answer: option A/ eibfjed

Suppose that the expected number of accidents per week at an industrial plant is 5. Suppose also that the number of workers injured in each accident are independent random variables with a common mean of 2.5. If the number of workers injured in each accident is independent of the number of accidents that occur, what is the expected number of workers injured in a week?

Answers

Answer:

12.5

Step-by-step explanation:

If the number of workers injured in each accident is independent of the number of accidents that occur, then the expected number of workers injured in a week equals the expected number of accidents per week times the expected number of workers injured in each accident.

But the expected number of workers injured in each accident is precisely the average or mean, so

Expected number of workers injured in a week = 5*2.5 = 12.5.

Or in other words, in 10 weeks there will probably be 125 workers injured.

Final answer:

The expected number of workers injured in a week at the industrial plant is 12.5, calculated by multiplying the expected number of accidents per week (5) by the expected number of workers injured per accident (2.5).

Explanation:

The expected number of accidents at an industrial plant per week is given as 5 and the mean number of workers injured per accident is given as 2.5. As the number of workers injured per accident is stated to be independent from the number of accidents that occur, we can merely multiply the two expected values to find the overall expected value of workers injured in a week.

So, Expected number of workers injured per week = Expected number of accidents per week * Expected number of workers injured per accident = 5 * 2.5 = 12.5.

This implies that, on average, we can expect 12.5 workers to be injured at the industrial plant per week.