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3 years ago

True or False if you took a true if then statement and inserted a not in each clause the new statement would also be true

Mathematics

1 answer:

nadya68 [22]3 years ago

8 0

Answer:

False

Step-by-step explanation:

Here's a true if-then statement:

if false, then true

Here's the modification proposed:

if not false, then not true . . . . . . . . . this is not a true if-then statement

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What is an example of a right angle, an acute triangle, and an obtuse triangle

Vitek1552 [10]

The example of a right angle is that it's exactly 90°, the acute angle is less than 90°, and the obtuse angle is more than 90° but less than 180°.

7 0

3 years ago

Identify each function that has a remainder of -3 when divided x+6

Sergeu [11.5K]

Answer:

Step-by-step explanation:

According to remainder theorem, you can know the remainder of these polynomials if you plug in x = -6 into them.

So we will plug in -6 into x of all the polynomials ( A through D) and see which one equals -3.

For A:

$x^5 + 2x^2 - 30x + 30\\=(-6)^5 + 2(-6)^2 - 30(-6) + 30\\=-7494$

For B:

$x^4 + 4x^3 - 21x^2 - 53x + 12\\=(-6)^4 + 4(-6)^3 - 21(-6)^2 - 53(-6) + 12\\=6$

For C:

$x^3 - 10x^2 - 7\\=(-6)^3 - 10(-6)^2 - 7\\=-583$

For D:

$x^4 + 6x^3 - 10x - 63\\=(-6)^4 + 6(-6)^3 - 10(-6) - 63\\=-3$

The only function that has a remainder of -3 when divided by x + 6 is the fourth one, answer choice D.

5 0

2 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,

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

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

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