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ale4655 [162]
2 years ago
7

The range of all real numbers x such that 2x − 5 < 7 OR 4x + 10 > 6 is___

Mathematics
1 answer:
scoray [572]2 years ago
4 0
First, simplify the two inequalities.

2x - 5 < 7
2x < 12 <-- Add 5 to each side
x < 6 <-- Divide both sides by 2

4x + 10 > 6
4x > -4 <-- Subtract both sides by 10
x > -1 <-- Divide both sides 4

Now we have the compound inequality x < 6 OR x > -1

To fulfill the first inequality, x has to be less 6. For the second inequality, x has to bigger than -1.

So, the answer is B. (-infinity, infinity).
You might be interested in
Your school organizes a clothing drive as a fundraiser for your class trip. The school earns $100 for every 400 pounds of donati
noname [10]

Answer:

$550

Step-by-step explanation:

400x=2200, x is the number of groups needed to get 2200 pounds.

If you divide both sides by 400, you'll get x=5.5

Next, multiply 5.5 by $100 because each group gives $100

hope this helps!

4 0
1 year ago
Read 2 more answers
Which classification best describes the following system of equations? 3x+6y-12z=36 x=2y-4z=12 4x+8y-16z=48 inconsistent and dep
Viktor [21]

<u>Answer:</u>

Consistent and dependent

<u>Step-by-step explanation:</u>

We are given the following equation:

1. 3x+6y-12z=36

2. x+2y-4z=12

3. 4x+8y-16z=48

For equation 1 and 3, if we take out the common factor (3 and 4 respectively) out of it then we are left with x+2y-4z=12 which is the same as the equation number 2.

There is at least one set of the values for the unknowns that satisfies every equation in the system and since there is one solution for each of these equations, this system of equations is consistent and dependent.

6 0
2 years ago
Tickets for the baseball games were $2.50 for general admission and 50 cents for kids. If there were six times as many general a
Makovka662 [10]
There was 3000 general admission tickets sold and 500 kid ticket sold.

How did I get this?

First, we need to see what information we have.
$2.50 = General admission tickets = (G)
 $0.50 = kids tickets =  (K)
There were 6x as many general admission tickets sold as kids. G = 6K

We need two equations:
G = 6K  
$2.50G + $.50K = $7750
Since, G = 6K we can substitute that into the 2nd equation.

2.50(6K) + .50K = 7750
Distribute 2.50 into the parenthesis

15K + .50K = 7750
combine like terms

15.50K = 7750
Divide both sides by 15.50, the left side will cancel out.

K = 7750/15.50
K = 500 tickets
So, 500 kid tickets were sold.

Plug K into our first equation (G = 6k)

G = 6*500
G = 3000 tickets

So, 3000 general admission tickets were sold,

Let's check this:

$2.50(3000 tickets) = $7500 (cost of general admission tickets)
$.50(500 tickets) = $250 (cost of general admission tickets)
$7500 + $250 = $7750 (total cost of tickets)




4 0
2 years ago
A local factory uses a manufacturing process in which 70% of the final products meet quality standards and 30% are found to be d
frosja888 [35]
<span>Answer:
   a manufacturing process has a 70% yield, meaning that 70% of the porducts are acceptable and 30% are defective. If three of the products are randomly selected, find the probability that all of them are acceptable. ---
   Ans: 0.7^3 = 0.3430.</span>
8 0
2 years ago
What is the following product? RootIndex 3 StartRoot 16 x Superscript 7 Baseline EndRoot times RootIndex 3 StartRoot 12 x Supers
KengaRu [80]

Answer:

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }

Step-by-step explanation:

Given

\sqrt[3]{16x^7} * \sqrt[3]{12x^9}

Required

Find the products

From laws of indices;

\sqrt[m]{a} * \sqrt[m]{b} = \sqrt[m]{a*b}

So;

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16x^7 * 12x^9}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16* x^7 * 12 * x^9}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^7  * x^9}

From laws of indices

a^m * a^n = a^(m+n); So,

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{7+9}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{16}}

Expand 16 * 12

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4*4*4*3* x^{16}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4^3 *3* x^{16}}

From laws of imdices

a^{\frac{1}{m}} = \sqrt[m]{a}

So;

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^3 *3* x^{16}})^{\frac{1}{3}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^{3*{\frac{1}{3}}} *3^{{\frac{1}{3}}}* x^{16*{\frac{1}{3}}}})

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{\frac{16}{3}}}

Divide 16 by 3 (Write as ,mixed number)

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5\frac{1}{3}}}

Split mixed numbers

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5+\frac{1}{3}}}

Apply law of indices

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5}*{x ^\frac{1}{3}}}

Reorder

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 * x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}

Apply law of indices

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3} *\sqrt[3]{x} }

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3*x} }

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3x} }

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }

6 0
2 years ago
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