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3 years ago

If AC = 10 and CG = 21, find AG.

Mathematics

1 answer:

stellarik [79]3 years ago

8 0

The answers 59 because 90 degrees is the angle so 21 plus 10 equals 31 and 90-31 equals 59

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I need the answer for number 2

Advocard [28]

<h3>I'll teach you how to solve -2/3 - 3/5</h3>

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-2/3 - 3/5

LCM of 3 and 5 is 15

Adjust fractions based on the LCM:

-10/15 - 9/15

Since the denominators are equal combine the fractions:

-10-9/15

Subtract the numbers:

-19/15

Apply fraction rule:

- 19/15

Your Answer Is The Second Option - 19/15

plz mark me as brainliest :)

7 0

3 years ago

The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate

KonstantinChe [14]

Answer:

It will take 47 hours

Step-by-step explanation:

We can have the exponential function as follows;

P = I( 1 + r)^n

if the initial value was x , the final will be 2x

The change rate is 1.5% , which is r, this is same as;

1.5/100 = 0.015

So what we want to calculate is the n attached

This will be;

2x = x ( 1 + 0.015)^n

2 = 1.015^n

ln 2 = n ln 1.015

n = ln 2/ln 1.015

n = 46.55

This is approximately 47 hours

7 0

2 years ago

Rationalize the numerator or denominator, and simplify:

allochka39001 [22]

$\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots$
$=\dfrac{x+h-x}\cdots=\dfrac h\cdots$

The $h$ s then cancel, leaving you with the $\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}$ term.

If it's not clear what I did above, consider the substitution $a=\sqrt[3]{x+h}$ and $b=\sqrt[3]x$ . Then

$a^3-b^3=(a-b)(a^2+ab+b^2)$
$\implies\dfrac{a-b}h=\dfrac{a-b}h\times\dfrac{a^2+ab+b^2}{a^2+ab+b^2}=\dfrac{a^3-b^3}{h(a^2+ab+b^2)}$