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Lana71 [14]
3 years ago
9

Determine which consecutive integers do not have a real zero of between them. a. (–6, –5) c. (–2, –1) b. (–5, –4) d. (1, 2) Plea

se select the best answer from the choices provided A B C D
Mathematics
1 answer:
svetoff [14.1K]3 years ago
7 0
We have by the intermediate value theorem that if a continuous function takes values both above and below zero at 2 points, there is a zero of the function in-between. We have that polynomials are continues. Let's calculate f(-6) and f(-5). f(-6)=-36 while f(-5)=-1. Thus, we cannot conclude that there is a root between them.
F(-2)=8, f(-1)=-1, so there is a flip; a zero must exist between them.
F(1)=-1, f(2)=20, so again there is a change of signs.
f(-5)=-1, f(-4)=14 so there is a root still.
We have that the only choice that does not have a root between the integers is choice a.

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<u>★</u><u> </u><u>Divisibility</u><u> </u><u>rule</u><u> </u><u>for</u><u> </u><u>3</u><u> </u><u>-</u>

If the sum of all the digits divisible by 3 then the number is divisible by 3.

Here, we take an example for better understanding -

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★<u> </u><u>Divisibility</u><u> rule</u><u> </u><u>for</u><u> </u><u>6</u><u> </u><u>-</u>

A number is to be divisible by 6 , if the number is also divisible by 2 and 3.

<u>Exampl</u><u>e</u><u> </u><u>2</u><u> </u><u>-</u> 132

Add up the digits = 1 + 3 + 2 = 6

As, 6 is divisible by 2 and it's also divisible by 3. Then, 132 is a factor of 6.

★ <u>Divisibility</u><u> rule</u><u> for</u><u> </u><u>9</u><u> </u><u>-</u>

If the sum of all the digits is divisible by 9 then the number is divisible by 9. Just like we'd in 3.

<u>Example</u><u> </u><u>3</u><u> </u><u>-</u> 1052

Add up the digits - 10 + 5 + 2 = 17

As, 17 is not divisible by 9. So, 1052 is also not divisible by 9.

3 0
3 years ago
Simplify 5^-2 x 5^5 x 5
tatiyna
5^-2 = .04
5^5 = 3125

.04 x 3125 x 5 = 625





4 0
3 years ago
Read 2 more answers
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