Evaluate the following function is f(x)=3x-2 and g(x)=5x+1 what is -2•f(4)+3•g(1)

Answers

Answer 1

Answer:

Answer:

work is shown and pictured

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The hardcover version of a book weighs twice as much as its paperback version. The hardcover book and the paperback together weigh 4.2 pounds. Which system of equations can be used to find h, the weight of the hardcover book, and p, the weight of the paperback?

Answers

If the weight of hardcover version is double the weight of paperback version then weight of hardcover version is 2.8 pounds and the weight of paperback version is 1.4 pounds.

What is equation?

An equation is a relationship between variables in equal to form. It is evaluated to find the values of variables.

How to solve an equation?

The weight of hardcover book is h

The weight of paperback version is p.

According to question:

h=2p (weight of hardcover version is double the weight of paperback version)

h+ p= 4.2 (weight of hardcover version and weight of paperback version is equal to 4.2 pounds)

Put the value of h=2p in h+ p=4.2

2p+p=4.2

3p=4.2

p=1.4

Put the value of p=1.4 in h=2p to get the value of h

h=2p

=2*1.4

=2.8

Hence the weight of hardcover version is 2.8 pounds and the weight of paperback version is 1.4 pounds.

Learn more about equations at brainly.com/question/2972832

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Answer:

Step-by-step explanation:

on edge 2023

A hospital employs a total of 77n nurses and doctors. The ratio of nurses to doctors is 9:2. How many nurses are employed at the hospital? How many doctors are employed at the hospital.

Answers

Answer:

number of nurses employed at the hospital is 9x = 9 * (7n/9) = 7n, and the number of doctors is 2x = 2 * (7n/9) = (14n/9).

Step-by-step explanation:

Given the ratio of nurses to doctors is 9:2, we can express this as:

Number of nurses (9x) : Number of doctors (2x)

Now, you've mentioned that the total number of nurses and doctors employed at the hospital is 77n. So, we can set up an equation:

9x + 2x = 77n

Combine like terms:

11x = 77n

Now, to find the number of nurses (9x) and the number of doctors (2x), we need to divide both sides by 11:

9x = (77n) / 11

9x = 7n

Now, we can solve for x:

x = (7n) / 9

So, the number of nurses employed at the hospital is 9x = 9 * (7n/9) = 7n, and the number of doctors is 2x = 2 * (7n/9) = (14n/9).

In summary:

- Number of nurses = 7n

- Number of doctors = (14n/9)

solved by MG

Use integration by parts to integrate sin2x between pi and 0

Answers

Answer:

$\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = 0$

General Formulas and Concepts:

Calculus

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 2x$
[u] Differentiate: $\displaystyle du = 2 \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 0 ,\ u = 2(0) = 0} \atop {x = \pi ,\ u = 2 \pi}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2) \int\limits^0_(\pi) {2 \sin (2x)} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2) \int\limits^0_(2 \pi) {\sin u} \, du$
Trigonometric Integration: $\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2)(-\cos u) \bigg| \limits^0_(2 \pi)$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = (1)/(2)(0)$
Simplify: $\displaystyle \int\limits^0_(\pi) {\sin (2x)} \, dx = 0$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

$25;300%increse percent change

Answers

well and 300% increase means you multiply by 3 and then add the original number

in this case 300% of 25 is 75 and its an increase so 25+75=100

the answer is 100

6(7a-10) how do I solve this

Answers

$6(7a-10) \n \n 6 * 7a + 6 * -10 \ / \ distribute \n \n 42a + 6 * -10 \ / \ simplify \n \n 42a + -60 \ / \ simplify \n \n 42a - 60 \ / \ simplify \n \n Answer: \fbox {42a - 60}$

solve the brackets, to do that you should multiply the numbers inside the bracket with the first number outside and in the right to the brackets.so,

6(7a-10)= 42a-60
hence the answer is, 42a-60

Suppose you throw a dart at a circular target of radius 10 inches. Assuming that you hit the target and that the coordinates of the outcomes are chosen at random, find the probability that the dart falls (a) within 2 inches of the center. (b) within 2 inches of the rim. (c) within the first quadrant of the target. (d) within the first quadrant and within 2 inches of the rim.

Answers

Answer:

a) The probability is 0.04

b) The probability is 0.36

c) The pprobability is 0,25

d) The probability is 0.09

Step-by-step explanation:

Lets calculate areas:

the target has a radius of 10 inces, hence the target area has a area on 10²*π = 100π square inches.

a) A circle of 2 inches of radius has an area of 2²π = 4π square inches, hence the probability of hitting that area is 4π/100π = 1/25 = 0.04

b) If the dart s within 2 inches of the rim, then it is not at distance 8 inches from the center (that is the complementary event). The probability for the dart to be at 8 inches of the center is 8²π/100π = 64/100 = 16/25 = 0.64, thus, the probability that the dart is at distance 2 or less from the rim is 1-0.64 = 0.36.

c) The first quadrant has an area exactly 4 times smaller than the area of the target (each quadrant has equal area), thus the probability for the dart to fall there is 1/4 = 0.25

d) If the dart is within 2 inches from the rim (which has probability 0.36 as we previously computed), then it will be equally likely for the dart to be in either of the 4 quadrants (the area that is within 2 inches from the rim forms a ring and it has equal area restricted on each quadrant). Therefore, the probability for the dice to be in the first qudrant and within 2 inches from the rim is 0.36*1/4 = 0.09.

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