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3 years ago

6 x - 2 [ x + 2 ] > 2 - 3 [ x + 3] could u show me step by step on how to do thiz

Mathematics

1 answer:

Dafna11 [192]3 years ago

8 0

Answer:

x > - 3/7

Step-by-step explanation:

6x - 2x - 4 > 2 - 3 (x + 3)

6x - 2x - 4 > 2 - 3x - 9

4x - 4 > 2 - 3x - 9

4x - 4 > - 7 - 3x

4x - 4 + 3x > - 7

7x > - 7 + 4

7x > - 3

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Hey,can you help me please i very need help..

Brut [27]

Answer:

4.71cm

Step-by-step explanation:

5 0

3 years ago

Write 90 as a product of prime factors in expanded form.

NISA [10]

Given :

The number is given as 90.

Explanation :

To find the product of prime factors in expanded form

$90=2\times3\times3\times5$

Answer :

Hence the answer is

$90=2\times3\times3\times5$

5 0

1 year ago

Given limit f(x) = 4 as x approaches 0. What is limit 1/4[f(x)]^4 as x approaches 0?

stepladder [879]

Answer:

$\displaystyle 64$

General Formulas and Concepts:

Calculus

Limits

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

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

Limit Property [Multiplied Constant]: $\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \lim_{x \to 0} f(x) = 4$

Step 2: Solve

Rewrite [Limit Property - Multiplied Constant]: $\displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4} \lim_{x \to 0} [f(x)]^4$
Evaluate limit [Limit Rule - Variable Direct Substitution Exponential]: $\displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4}(4^4)$
Simplify: $\displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = 64$

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

Unit: Limits

Book: College Calculus 10e

3 0

3 years ago

What is arccos(−1) in Radians.

Free_Kalibri [48]

Answer:

π radians

Step-by-step explanation:

The arc cos(- 1) in degrees is 180° and 180° = π radians

7 0

4 years ago

PLEASE HELP ME PLEASE!!

Harman [31]

Answer:

hypotenuse (h)=?

perpendicular (p)=3

base(b)=2

using pythagoras theorem

h²=p²+b²

h²=3²+2²

h²=9+4

h²=13

h=√13

hypotenuse (h)=12
perpendicular (p)=10
base(b)=?

using pythagoras theorem

b²=h²-p²

b²=12²-10²

b²=144-100

b=√44

b=2√11

6 0

3 years ago