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3 years ago

Someone please help and show work!! I will also mark brainliest. I’ll appreciate it.

Mathematics

1 answer:

Eduardwww [97]3 years ago

5 0

Answer:

1. 53

2. 20

3. 37

4. 29

Step-by-step explanation:

Since your trying to find the hypotenuse you equation is...

1. 28^2 + 45^2 = C^2

784 + 2025 = C^2

Square Root 2809= C^2

C=53

So after you square root 2809 you get 53.

2. 12^2 + 16^2 = C^2

144 + 256 = C^2

Square Root 400 = C^2

C=20

So after you square root 400 you get 40.

3. 12^2 + 35^2 = C^2

144 + 1225 = C^2

Square Root 1369 = C^2

C=37

So after you square root 1359 you get 37.

4. 20^2 + 21^2 = C^2

400 + 441 = C^2

Square root 841 = C^2

C=20

So after you square root 841 you get 29.

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Y=(4x-5)(x-?) what does x equal

Whitepunk [10]

Set the factor '(4 + -1x)' equal to zero and attempt to solve: Simplifying 4 + -1x = 0 Solving 4 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-4' to each side of the equation. 4 + -4 + -1x = 0 + -4 Combine like terms: 4 + -4 = 0 0 + -1x = 0 + -4 -1x = 0 + -4 Combine like terms: 0 + -4 = -4 -1x = -4 Divide each side by '-1'. x = 4 Simplifying x = 4

Easy answer
Its 0

I hope it's correct Good luck...

8 0

3 years ago

Urgent. Please show all work

myrzilka [38]

Answer:

$\displaystyle f'(x) = \frac{4}{x^2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Differentiation

Derivatives
Derivative Notation

The definition of a derivative is the slope of the tangent line: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle f(x) = -\frac{4}{x}$

Step 2: Differentiate

[Function] Substitute in x: $\displaystyle f(x + h) = -\frac{4}{x + h}$
Substitute in functions [Definition of a Derivative]: $\displaystyle f'(x) = \lim_{h \to 0} \frac{-\frac{4}{x + h} - \big( -\frac{4}{x} \big)}{h}$
Simplify: $\displaystyle f'(x) = \lim_{h \to 0} \frac{4}{x(x+ h)}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle f'(x) = \frac{4}{x(x+ 0)}$
Simplify: $\displaystyle f'(x) = \frac{4}{x^2}$