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2 years ago

PLEASE NO LINKS OR SITES!!!! JUST ANSWERS. THANK YOU!

Mathematics

1 answer:

lutik1710 [3]2 years ago

6 0

Answer:

F=-9

Hope This Helps!

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Which equation in point-slope form contains the point (–3, 5) and has slope –1? y + 3 = –1(x – 5) y – 5 = –1(x + 3) y – 3 = –1(x

andreev551 [17]

Answer:

y - 5 = -1(x + 3)

Step-by-step explanation:

Point slope form

y - y1 = m(x - x1) where m = slope and passing through point (x1 , y1)

In this case if the equation in point-slope form contains the point (–3, 5) and has slope –1 then the equation should be:

y - 5 = -1(x - (-3))

y - 5 = -1(x + 3)

5 0

3 years ago

Write an equation in point slope form of the he that passes through the point (2.1) and has a slope of 2

Wewaii [24]

Answer:

y=2x-3

Step-by-step explanation:

show work

1=(2*2)+b

1=4+b

1-4=-3

-3=b

check work

y=2x-3

y=(2*2)-3

y=4-3

y=1

5 0

3 years ago

Read 2 more answers

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.

7 0

3 years ago

Emily gad a decorative box that is shaped like a cube with a height of 5 inches. What’s the surface area of the box?

lina2011 [118]

times the length by width and you will get your answer use a calculator that will be the answer

4 0

3 years ago

Janice needs 3 gallons of lemonade for a party. She has 4 quarts, 6 pints, and 4 cups of lemonade already made. How much more le

Allisa [31]

4 quarts is 1 gallon
6 pints is 0.75 gallons
4 cups is 0.25 gallons
if you add all of those you would get 2 gallons
so she would need 1 more gallon of lemonade

7 0

3 years ago