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3 years ago

Can anyone tell the answer

Mathematics

1 answer:

QveST [7]3 years ago

6 0

First of all, as the line (y<3x) is dotted, so the sign of its inequality is just <, not <=

Options b ad c are out

To check which region is the right one for the second inequality (y>=-0.5x-4), plug in the origin point (0,0) into the inequality, if it satisfies then the region included the origin is the shaded region (the solution you want), if not, the other region is the solution..

0>+0-4 0>=-4

Yes, it satisfies,,

SO it is A

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7 0

3 years ago

Simplify 3 - (4x - 5) + 6. A) 12x + 21 B) 4x + 14 C) 4x + 4 D) -4x + 14

stepladder [879]

Answer:

D) -4x+14

Step-by-step explanation:

We are given with the expression

3 - (4x - 5) + 6

Here we are going to open the bracket at first step

as we have a negative sign in front of bracket the signs of the terms inside brackets will be reversed

3-4x+5+6

now we bring the like terms together

3+5+6 - 4x

Here 3, 5, 6 all are constant and hence like

14-4x

3 - (4x - 5) + 6= -4x+14

Hence our answer is -4x+14

6 0

3 years ago

At a school there are 864 students and 80 staff members. 5/8 of the students are boys. How many girls are in the school?

alexgriva [62]

Answer:

354

Step-by-step explanation:

3/8 multiplied by 864

7 0

3 years ago

Read 2 more answers

How to do long division with polynomials?

myrzilka [38]

Divide the first term of the numerator by the first term of the denominator, and put that in the answer.Multiply the denominator by that answer, put that below the numerator.Subtract to create a new polynomial.

Hope this helps. :)

5 0

3 years ago

How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi

ryzh [129]

Answer:

3003 number of 5-member chess teams can be chosen from 15 interested players.

Step-by-step explanation:

Given:

Number of the interested players = 15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula $nCr = \frac{n!}{r!(n - r)!}$

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now we have n = 15 and r = 5

Substituting the values,

$15C_5 = \frac{15!}{5!(15- 5)!}$

$15C_5 = \frac{15!}{5!(10)!}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11}{(5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{360360}{120}$

$15C_5 = 3003$

7 0

3 years ago