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3 years ago

(ii) n is an integer and 6n +5 <20

Mathematics

1 answer:

Sonja [21]3 years ago

3 0

6n + 5 < 20

Subtract both sides 5

6n + 5 - 5 < 20 - 5

6n < 15

Divide both sides by 6

6n ÷ 6 < 15 ÷ 6

n < 15/6

n < 3×5/3×2

n < 5/2

n < 2.5

n《 2

Thus the largest value of n is :

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How to solve f(4) = x^2+7x-22

geniusboy [140]

Answer:

Step-by-step explanation:

1. Plugin 4 for each corresponding X value

2. Should look like 4^2+7(4)-22

3. Once ran through a calculator you should end up with a value of 22

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2 years ago

Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con

monitta

Let $f(x)=e^{-1/x}$ . Then $f'(x)=\frac1{x^2}e^{-1/x}>0$ for all $x\ge2$ , so $f$ is strictly increasing. As $x\to\infty$ , $e^{-1/x}\to e^0=1$ , so $f$ is bounded above by 1. This is to say,

$e^{-1/x}$

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We have, by setting $y=-\frac1x$ ,

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3 years ago

Math Story problem. I need help

Olin [163]

Answer:

i believe it would be 12

hope this helps !!

Step-by-step explanation:

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3 years ago

Anybody know the anwser

Darya [45]

Answer:

A 3 goes in the parentheses. The expression evaluates to 59.

Step-by-step explanation:

b = 3

2b^3 + 5 =

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3 years ago

HELP!!!!

ch4aika [34]

Answer:

$8 y^{50}$

Step-by-step explanation:

Given (64 y Superscript 100 Baseline) Superscript one-half.

Let us write it into an equation.

$\left(64 y^{100}\right)^{\frac{1}{2}}$

Apply radical rule: $\sqrt[n]{a}=a^{\frac{1}{2}}$ and $a^{m+n}=a^m+a^n$

$\begin{aligned}\left(64 y^{100}\right)^{\frac{1}{2}} &=\sqrt[2]{64 y^{100}} \\&=\sqrt[2]{8^{2} y^{50} y^{50}} \\&=\sqrt[2]{8^{2}\left(y^{50}\right)^{2}} \\&=8 y^{50}\end{aligned}$

Hence, $8 y^{50}$ is equivalent to (64 y Superscript 100 Baseline) Superscript one-half.

8 0

3 years ago