All categories

3 years ago

Which of the following are coefficient

Mathematics

1 answer:

malfutka [58]3 years ago

4 0

Answer:

2 and -3

Step-by-step explanation:

because a coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4x).

so in 2x-3y+5, 2 and -3 are the coefficients.

You might be interested in

115+150+150+150+150-100

tresset_1 [31]

Answer 415

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

Find the equivalent expression of 3+2(2p+4t)

NISA [10]

Remember to follow PEMDAS, and to only combine terms with like variables.

First, distribute the 2 to all terms within the parenthesis:

2(2p + 4t) = 2(2p) + 2(4t) = 4p + 8t

3 + 4p + 8t, or (D), is your answer choice, for it cannot be simplified anymore.

7 0

4 years ago

Allyson and Abby were throwing the football. Allyson threw it 38 feet and Abby threw it 46 feet. How many YARDS did they throw i

viva [34]

Answer:

28 yards

Step-by-step explanation:

38+46=84

1 yard = 3 feet

1/3=x/84

84/3=24

5 0

3 years ago

The profit earned by a hot dog stand is a linear funsion of the number of hot dogs sold.It cost the owner $48 dollars each morni

Delvig [45]

Answer:

Step-by-step explanation:

If he earns $2 per hot dog sold, then the algebraic expression for this is 2x, where x is the number of hot dogs sold. If he has to put out $48 to buy the supplies, that is represented as -48. The function, then, that represents the profit earned by selling the hot dogs is

y = 2x - 48

5 0

3 years ago

In order to conduct an experiment, 4 subjects are randomly selected from a group of 20 subjects. How many different groups of fo

irga5000 [103]

Answer:

The number of ways to form different groups of four subjects is 4845.

Step-by-step explanation:

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!\times (n-k)!}$

In this case, 4 subjects are randomly selected from a group of 20 subjects.

Compute the number of ways to form different groups of four subjects as follows:

${n\choose k}=\frac{n!}{k!\times (n-k)!}$

${20\choose 4}=\frac{20!}{4!\times (20-4)!}$

$=\frac{20\times 19\times 18\times 17\times 16!}{4!\times 16!}\\\\=\frac{20\times 19\times 18\times 17}{4\times3\times 2\times 1}\\\\=4845$

Thus, the number of ways to form different groups of four subjects is 4845.

5 0

3 years ago