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3 years ago

-4x+9(10x-1) please help me this is distributive property

Mathematics

1 answer:

kotykmax [81]3 years ago

3 0

Let me walk you through this real quick:

-4x + 9 (10x - 1)
-4x + 90x - 9 (open up the brackets)
90x - 4x - 9
= 86x - 9

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Help Me With These Please

anzhelika [568]

10×sine of 60 degrees=8.66 which is the same as B. you can check by finding the square root of 3 and then multiplying that by 5.

6 0

3 years ago

Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997, However in march 1997 the price of the tea increased by 25% and

Irina18 [472]

Answer: tea = 15 rupees per kg

sugar= 3 rupees per kg

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations with the information given:

"Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997":

2 t + 3 s =39 (a)

Where:

t= price of 1 kg of tea
s = price of 1 kg of sugar

"in march 1997 the price of the tea increased by 25% (1.25)and the price of the sugar increased by 20%(1.20) and the same quantity of tea and sugar cost rupees 48.30. "

2(t1.25)+3(s1.2) = 48.30 (b)

Solving for t in (b)

2t =39-3s

t = (39 -3s)/2

t = 19.5-1.5s

Replacing the value of t in (b)

2 x ((19.5-1.5s)1.25)+ 3 ( 1.2s) =48.30

2x ( 24.375 -1.875s) +3.6s =48.30

48.75 -3.75s+3.6s= 48.30

48.75-48.30 = 3.75s-3.6s

0.45= 0.15s

0.45/0.15 =s

3 =s

Replacing the value of s in (a)

2 t + 3 (3) =39

2 t + 9 =39

2 t =39 -9

2 t =30

t = 30/2

t= 15

Prices in january:

tea = 15 rupees per kg

sugar= 3 rupees per kg

Feel free to ask for more if needed or if you did not understand something.

4 0

3 years ago

Read 2 more answers

PLEASE HELP ME GUYS OR I WONT PASS this calculus!!!!

KonstantinChe [14]

Answer:

b. $\displaystyle \frac{1}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation
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

$\displaystyle H(x) = \sqrt[3]{F(x)}$

Step 2: Differentiate

Rewrite function [Exponential Rule - Root Rewrite]: $\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}$
Chain Rule: $\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]$
Basic Power Rule: $\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)$
Simplify: $\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}$
Substitute in function values: $\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}$
Exponents: $\displaystyle H'(5) = \frac{6}{3(4)}$
Multiply: $\displaystyle H'(5) = \frac{6}{12}$
Simplify: $\displaystyle H'(5) = \frac{1}{2}$