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VikaD [51]
3 years ago
15

The force of an object (f) is equal to its mass (m) times its acceleration (a). Thus, the formula for calculating force is f = m

a. 
Based on this information, which of the following equations would be used to calculate acceleration? 

       A. a = fm   B. a = f + m   C. a = f − m   D. a = f ÷ m
Mathematics
2 answers:
Juliette [100K]3 years ago
6 0

Answer:

D. a=f\div m.

Step-by-step explanation:

We are given that force of an object and we have to find the equation would be used to calculate the acceleration

Mass of object=m

Acceleration of an object= a

Force of an object=f

We are given that force of an object is equal to its mass times its acceleration.

Therefore , the formula for calculating force

f=ma

Because we are given this formula

Now , we have find the equation which is used to calculate  acceleration

Then by division property of equality we have

a=f\div m

Hence, the equation would be used to calculate acceleration a=f\div m.Therefore, option D is true.

Grace [21]3 years ago
5 0
Your answer would be D. Bc you haven't divide a from b to get f on the other side if the equation.
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3 years ago
Read 2 more answers
Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America
Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

   P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a  student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is  written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

4 0
3 years ago
Verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial
Mariulka [41]

Answer:

i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

ii) - the sum of the zeros and the corresponding coefficients are the same

-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.

-the product of the zeros of the polynomial is same as the corresponding coefficient

Step-by-step explanation:

We are given the cubic polynomial;

p(x) = 2x³ - 3x² - 3x + 2

For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.

Thus;

P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0

P(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0

P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0

Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

Now, let's verify the relationship between the zeros and the coefficients.

Let the zeros be as follows;

α = 2

β = -1

γ = ½

The coefficients are;

a = 2

b = -3

c = -3

d = 2

So, the relationships are;

α + β + γ = -b/a

αβ + βγ + γα = c/a

αβγ = -d/a

Thus,

First relationship α + β + γ = -b/a gives;

2 - 1 + ½ = -(-3/2)

1½ = 3/2

3/2 = 3/2

LHS = RHS; So, the sum of the zeros and the coefficients are the same

For the second relationship, αβ + βγ + γα = c/a it gives;

2(-1) + (-1)(½) + (½)(2) = -3/2

-2 - 1½ + 1 = -3/2

-1½ - 1½ = -3/2

-3/2 = - 3/2

LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient

For the third relationship, αβγ = -d/a gives;

2 * -1 * ½ = -2/2

-1 = - 1

LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient

7 0
3 years ago
I don’t understand need help
RideAnS [48]

Answer: Choice D

(a-e)/f

=======================================

Explanation:

Points D and B are at locations (e,f) and (a,0) respectively.

Find the slope of line DB to get

m = (y2-y1)/(x2-x1)

m = (0-f)/(a-e)

m = -f/(a-e)

This is the slope of line DB. We want the perpendicular slope to this line. So we'll flip the fraction to get -(a-e)/f and then flip the sign from negative to positive. That leads to the final answer (a-e)/f.

Another example would be an original slope of -2/5 has a perpendicular slope of 5/2. Notice how the two slopes -2/5 and 5/2 multiply to -1. This is true of any pair of perpendicular lines where neither line is vertical.

8 0
2 years ago
Find the x and y intercepts for the following equation. 2x– 7y=-14
Kisachek [45]

Answer:

C: Xint = 2, Yint. = -7

Step-by-step explanation:

Given the equation, 2x - 7y = -14:

The y-intercept is the point on the graph where it crosses the y-axis, and has coordinates (0, b). It is also the value of y when x = 0.  To solve for the y-intercept, set x = 0:

2x - 7y = -14

2(0) - 7y = -14

0 - 7y = -14

-7y = -14

Divide both sides by -7 to solve for y:

-7y/-7 = -14/-7

y = 2

Therefore, the y-intercept = 2.

Next, the x-intercept is the point on the graph where it crosses the x-axis, and has coordinates, (a, 0). To find the x-intercept, set y = 0 and substitute into the given equation:

2x - 7y = -14

2x - 7(0) = -14

2x - 0 = -14

2x = -14

Divide both sides by 2 to solve for x:

2x/2 = -14/2

x = -7

x-intercept = -7.

Therefore, the correct answer is C: Xint = 2, Yint. = -7

Please mark my answers as the Brainliest if you find this helpful :)

7 0
2 years ago
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