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3 years ago

Does somene know easy if u know it

Mathematics

2 answers:

Levart [38]3 years ago

8 0

Answer:

valid

Step-by-step explanation:

because area of a square if found by A=a^2

(this can also be length times height since a square has all 4 equal sides)

3*3=9

Nookie1986 [14]3 years ago

4 0

Answer:

valid bc in a square all the sides are the same.

Step-by-step explanation:

so side 1 and side 2 both equal 3. 3×3=9

so the answer is valid

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In the number 432.785 which number is the hundredths digit

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The number in the hundredths digit is 8.

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joe is in debt $250, but he's saving $30 each month. using the equation, y=30x-250, how long will it take joe to have $230?

Vsevolod [243]

Answer:

Step-by-step explanation:

so in this equation we have 2 variables, and we must figure out what those variables mean:

y: "y" is the total amount of money joe has after subtracting his debt from it

x: "x" is the # of months as the cost of each month(30) is placed just before this variable

as y is the total amount of money joe has we need to substitute it for 230 to figure out how long it took for joe to recieve that much money

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3 years ago

At Norman's Newsstand, 5 magazines cost $10.00. How many magazines could you buy with $24.00?

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Step-by-step explanation:

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Determine whether each equation is a linear equation. Write yes or no. If yes, write the equation in standard form. 12x = 7y - 1

Anna71 [15]

Answer:

Yes, this is linear.

y = -4x

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2 years ago

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Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

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3 years ago