All categories

3 years ago

Please find Area of triangle

Mathematics

1 answer:

horrorfan [7]3 years ago

4 0

Answer:

Step-by-step explanation:

Side A would end up being 15 and then to find the area you would use your formula 1/2 × base × height.

You might be interested in

Find FL if H is the circumcenter of EFG,

Zanzabum

The artistic crop isn't helpful; it cuts off some vertex names.

The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so

EH² = HL² + EL²

EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339

Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.

Answer: 4.25

8 0

3 years ago

HELP ASAP!! BRAINLIEST

Maslowich

Answer: red is exponential growth blue is exponential decay and green is neither

Step-by-step explanation:

5 0

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