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3 years ago

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Help im suck at math pleasee 1. simplify 2.9+0.3-12-6 2. simplify 2.9-5.3-4n+6

1 answer:

ziro4ka [17]3 years ago

8 0

Answer:

n=0.956

Step-by-step explanation:

2.9+6=8.9 and 5.3n - 4n = -9.3n so, 8.9 - 9.3n = ? so -9.3n = -8.9 and 0.3n = 8.9 so n = 8.9/9.3 which is 0.956

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Graph the solution set for the inequalíty

marissa [1.9K]

Answer:

Slope= 2

Y-intercept= -5

Step-by-step explanation:

This inequality is in slope-intercept form which means we can get both striaght from the inequaltiy. Slope-intercept form is y=mx+b, where m is the slope and b is the y-intercept. In this inequality m would be 2 and b would be -5, so those are your answers.

5 0

3 years ago

PLS HELP ITS URGENT!!

DiKsa [7]

Answer:

SA= 28 in^2

V= 14 in^3

Step-by-step explanation:

circumference= 2πr

9.4 = 2πr

r = 1.496

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3 years ago

A cinema can hold 270 people at one performance 5/9 of the seats were occupied of the occupied seats 40% we occupied by concessi

Elza [17]

Question:

A cinema can hold 270 people at one performance 5/9 of the seats were occupied of the occupied seats 40% we occupied by concessionary ticket holders.

What is the number of seats occupied by concessionary ticket holders?.

Answer:

60 seats

Step-by-step explanation:

Given

Number of seats = 270

Occupied Seats = 5/9

Concessionary ticket holders = 40% of occupied Seats

Required

The number of seats occupied by concessionary ticket holders

First the number of occupied seat has to be calculated.

$Occupied\ Seats = \frac{5}{9} * 270$

$Occupied\ Seats = \frac{1350}{9}$

$Occupied\ Seats = 150$

Next is to determine the number of seats occupied by concessionary ticket holders.

$Number = 40\%\ of\ occupied\ seats$

$Number = 40\%\ of\ 150$

Convert percentage to decimal

$Number = 0.4 * 150$

$Number = 60$

<em>Hence, 60 seats were occupied by concessionary ticket holders.</em>

5 0

3 years ago

write an algebraic expression for this word problem. An orange contains 62 calories which is x fewer calories than a nectarine.

nasty-shy [4]

Answer:

n = 62 + x

Step-by-step explanation:

From the problem given, it is clear that the orange has LESS calories than a nectarine. <em>HOW much less? "x" less.</em>

<em />

Also, orange has 62 calories. So, nectarine MUST HAVE 62 + x calories.

<em>If we let nectarine's calories be n, we can write:</em>

<em>n = 62 + x</em>

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3 years ago

What is the roots of f(x)=-x^2-6x-14

Nutka1998 [239]

Answer:

$x=-3\pm i \sqrt{5}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS
Equality Properties

<u>Algebra I</u>

Standard Form: ax² + bx + c = 0
Quadratic Formula: $x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

<u>Algebra II</u>

Imaginary roots: √-1 = i

Step-by-step explanation:

<u>Step 1: Define function</u>

f(x) = -x² - 6x - 14

<u>Step 2: Set up</u>

Set equation equal to 0: -x² - 6x - 14 = 0
Factor out -1: -(x² + 6x + 14) = 0
Divide both sides by -1: x² + 6x + 14 = 0

<u>Step 3: Define variables</u>

a = 1

b = 6

c = 14

<u>Step 4: Find roots</u>

Substitute: $x=\frac{-6\pm\sqrt{6^2-4(1)(14)} }{2(1)}$
Exponents: $x=\frac{-6\pm\sqrt{36-4(1)(14)} }{2(1)}$
Multiply: $x=\frac{-6\pm\sqrt{36-56} }{2}$
Subtract: $x=\frac{-6\pm\sqrt{-20} }{2}$
Factor: $x=\frac{-6\pm\sqrt{-1} \sqrt{20} }{2}$
Simplify: $x=\frac{-6\pm2i \sqrt{5} }{2}$
Factor: $x=\frac{2(-3\pm i \sqrt{5} )}{2}$
Divide: $x=-3\pm i \sqrt{5}$

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3 years ago