Answer:
c is the correct option
Step-by-step explanation:
from,
f'(x) = h >0 <u>f</u><u>(</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u>
h
f(x) = - √2x
f(x + h) = - √(2x + h)
f'(x) = h>0 <u>-</u><u>√(2x + h) - √2x</u>
h
rationalize the denominator
= h>0 <u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>+</u><u> </u><u>√</u><u>2</u><u>x</u><u> </u><u> </u><u>(</u><u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>√</u><u>2</u><u>x</u><u>)</u>
h (-√(2x + h) - √2x)
= h>0 <u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>2</u><u>h</u><u> </u><u>-</u><u> </u><u>4</u><u>x</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
h(-√(2x + h) -√2x)
= h>0 <u>2</u><u>h</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
h(-√(2x+h) - √2x)
= h>0 <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
-√(2x+h) - √2x
∫ e^(3x)*(cosh(2x)dx
= ∫ [e^(3x)*(e^(2x)+e^(-2x))/2]dx
= ∫ [(e^(5x)+e^x)/2]dx
=e^(5x)/10+e^x/2+C
=(1/10)(e^(5x)+5e^x)+C
Answer:

Step-by-step explanation:
Since this is a right triangle, we can use the Pythagorean Theorem to solve for the sides.

where <em>a</em> and<em> b</em> are the legs and <em>c</em> is the hypotenuse. In this triangle, we know the legs are 9 centimeters and 11 centimeters, or:
Substitute these values into the formula.

Solve the exponents.
- (9 cm)²= 9 cm*9 cm=81 cm²

- (11 cm)²= 11 cm*11 cm= 121 cm²

Add the values on the left side.

Since we are solving for c, we must isolate the variable. It is being squared and the inverse of a square is the square root. Take the square root of both sides.



We are told to round to the nearest tenth.
The 1 in the hundredth place tells us to leave the 2 in the tenth place.

The hypotenuse is equal to <u>14.2 centimeters.</u>
Answer:
<h2>No,they are not</h2>
Step-by-step explanation:
An algebraic expression must contain and unknown value to solve for.
Answer:
x = 32/7, or 4 4/7 meals
Step-by-step explanation:
Write and solve an equation of ratios, as follows:
7/8 lb 4 meals
------------ -------------
1 lb x
x will be the number of meals necessary for the dog to eat 1 lb of food.
Cross-multiplying, (7/8)x = 4.
Multiplying both sides by 8/7 is an easy way to isolate x:
(8/7)(7/8)x = (8/7)(4)
or
x = 32/7, or 4 4/7 meals