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3 years ago

How many y's are in this term "3y"? *

Mathematics

1 answer:

german3 years ago

4 0

Answer:

Step-by-step explanation:

cause 3y is 3 times y and that's just 3 times y and NOT y time y times y

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Number 18 pls help me

Harman [31]

Lila did it correctly. The answer is 324

Following PEMDAS, we first focus on the parenthesis. So we simplify 9-3 to get 6

So we go from
18*4^2+(9-3)^2
to
18*4^2+6^2

The next step is applying exponents. In this case, squaring the terms, so we go from
18*4^2+6^2
to
18*16+36

Next is multiplying
18*16+36
turns into
288+36

Finally, add up 288 and 36 to get 288+36 = 324

That confirms that Lila is correct

----------------------

The error that Rob made is that he computed 18*4^2+9^2-3^2 but it is NOT correct. Saying (x-y)^2 = x^2-y^2 isn't a true equation for all x and y. Again you have to simplify what is in the parenthesis first, and then you can square it. Or you must use the FOIL rule to expand out (9-3)^2

7 0

3 years ago

Find the missing side length of the right triangle.

mart [117]

Answer: 80.42

Step-by-step explanation:

3 0

3 years ago

Rachel use 2/3 of a package of cornbread mix she will use equal parts of the leftover mix to make two batches of cornbread what

kramer

0.5/3 or 1/6 per batch

5 0

3 years ago

Find the equation of the tangent line to the curve when x has the given value.

Leviafan [203]

Answer:

14) The equation of the tangent line to the curve $f(x)=-2-x^2$ at x = -1 is $y=2x-1$

15) The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

Step-by-step explanation:

14) To find the equation of a tangent line to a curve at an indicated point you must:

1. Find the first derivative of f(x)

$f(x)=-2-x^2\\\\\frac{d}{dx} f(x)=\frac{d}{dx}(-2-x^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx} f(x)=\frac{d}{dx}\left(-2\right)-\frac{d}{dx}\left(x^2\right)\\\\f'(x)=-2x$

2. Plug x value of the indicated point, x = -1, into f '(x) to find the slope at x.

$f'(-1)=-2(-1)=2$

3. Plug x value into f(x) to find the y coordinate of the tangent point

$f(-1)=-2-(-1)^2=-3$

4. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line

$y-y_1=m(x-x_1)\\y+3=2(x+1)\\y=2x-1$

5. Graph your function and the equation of the tangent line to check the results.

15) To find the rate of learning at the end of eight hours of instruction you must:

1. Find the first derivative of f(x)

$w(t)=15\sqrt[3]{t^2} \\\\\frac{d}{dt}w= \frac{d}{dt}(15\sqrt[3]{t^2})\\\\w'(t)=15\frac{d}{dt}\left(\sqrt[3]{t^2}\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt[3]{u},\:\:u=\left(t^2\right)\\\\w'(t)=15\frac{d}{du}\left(\sqrt[3]{u}\right)\frac{d}{dt}\left(t^2\right)\\\\w'(t)=15\cdot \frac{1}{3u^{\frac{2}{3}}}\cdot \:2t\\\\\mathrm{Substitute\:back}\:u=\left(t^2\right)$

$w'(t)=15\cdot \frac{1}{3(t^2)^{\frac{2}{3}}}\cdot \:2t\\w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}$

2. Evaluate the derivative a t = 8

$w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}\\\\w'(8)=\frac{10\cdot 8}{\left(8^2\right)^{\frac{2}{3}}}\\\\=\frac{80}{\left(8^2\right)^{\frac{2}{3}}}\\\\\left(8^2\right)^{\frac{2}{3}}=16\\\\=\frac{80}{16}\\\\w'(8) = 5 \frac{items}{hour}$

The rate of learning at the end of eight hours of instruction is $w'(8) = 5 \frac{items}{hour}$

7 0

4 years ago

If b+c-a/a, c+a-b/CA are in A.P., prove that a, b, c are in<br>Н.Р.

SOVA2 [1]

Answer:I love thee fact that I am a straight a Student but I don't wanna help you dumb people

Step-by-step explanation:

3 0

3 years ago