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2 years ago

B is the midpoint of AC and D is the mid point of CE solve for x given BD=3x + 5 and AE= 4x+20

Mathematics

1 answer:

Elena L [17]2 years ago

8 0

Answer:triangle BCD is similar to triangle ACE (side, angle C, side)

AC = 2 BC

therefore AE = 2 BD

therefore

4x+20 = 2 (3x+5) solve that

Step-by-step explanation:

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Find the inverse of the given relation.

il63 [147K]

Answer:

The answer is B.

Step-by-step explanation:

To obtain the inverse of a relation you simply interchange the x and y coordinates of each pair.

Here:

(3,1) becomes (1,3)

(3,1) becomes ((1,3)

(7, -7) becomes (-7,7) and

(12, -15) becomes (-15, 12)

6 0

3 years ago

a grocery store has 5 pounds of granola. One customer buys 2/3 pounds of granola and another buys 5/6 pounds. after these purcha

Pavlova-9 [17]

There would be 3 and 1/2 pounds left.

2/3 x 2 = 4/6
5/6
5/6 + 4/6 = 6/6 (1) and 3/6 (1/2)

3 0

2 years ago

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Use mental math to find the amount of a 12% shipping fee for an item that costs $250. Explain your steps.

makvit [3.9K]

The shipping fee is $30
1. 12%×250
= 250×12/100
= 2.5 ×12
= 5/2×12
=60/2
=30

8 0

3 years ago

Read 2 more answers

HELP ASAP I GIVE BIG BRAIN

sukhopar [10]

A function is a relation where there is only one output for every input. In other words for every x value, there is only one y value.

Example : Multiply by 3 is a very simple function.

Input : 0 Output : 0

Input : 1 Output : 3

Input : 2 Output : 6

Input : 3 Output :9

And so on!

Hope this helps!

6 0

2 years ago

Find <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20" id="TexFormula1" title=" \frac{dy}{dx} " alt=" \frac{d

nataly862011 [7]

Answer:

$\displaystyle y' = 2x + 3\sqrt{x} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = (x + \sqrt{x})^2$

Step 2: Differentiate

Chain Rule: $\displaystyle y' = 2(x + \sqrt{x})^{2 - 1} \cdot \frac{d}{dx}[x + \sqrt{x}]$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = 2(x + x^{\frac{1}{2}})^{2 - 1} \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]$
Simplify: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]$
Basic Power Rule: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 \cdot x^{1 - 1} + \frac{1}{2}x^{\frac{1}{2} - 1})$
Simplify: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2}x^{-\frac{1}{2}})$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2x^{\frac{1}{2}}})$
Multiply: $\displaystyle y' = 2[(x + x^{\frac{1}{2}}) + \frac{x + x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}]$
[Brackets] Add: $\displaystyle y' = 2(\frac{2x + 3x^{\frac{1}{2}} + 1}{2})$
Multiply: $\displaystyle y' = 2x + 3x^{\frac{1}{2}} + 1$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = 2x + 3\sqrt{x} + 1$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

4 0

2 years ago