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Leni [432]
3 years ago
7

Write each number as a product of primes 1. 120 2. 160

Mathematics
2 answers:
AleksAgata [21]3 years ago
5 0

Answer:

120=2*2*2*3*5

160=2*2*2*2*2*5

Step-by-step explanation:

Vlada [557]3 years ago
3 0

Answer:

Below in bold.

Step-by-step explanation:

1.   120 = 12  * 10

= 2*2*3*2*5

= 2*2*2*3*5

2.  160 = 10 * 16

= 2*5*2*2*2*2

= 2*2*2*2*2*5

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I DONT GET THIS: What is the Domain and Range of the function in context?
Anettt [7]

Answer:

c

Step-by-step explanation:

the domain is every possible input value and the range is every possible output value. since the outputs are all positive, the range is equal to all real numbers. and since the graph starts at 1,000, the domain is greater than or equal to 1,000

6 0
2 years ago
9% of 750............................
babunello [35]
67.5 is your answer.
8 0
3 years ago
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Not sure what to do here can someone please help
kumpel [21]

Answer:

  x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

__

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

  \sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

  x = -1: √(-1+2) +1 = √(3(-1)+3)   ⇒   1+1 = 0 . . . . not true

  x = 2: √(2+2) +1 = √(3(2) +3)   ⇒   2 +1 = 3 . . . . true . . . x = 2 is the solution

__

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

  u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}

Solutions to this equation are ...

  u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

  x = u² -2 = 2² -2

  x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0
2 years ago
4. Six numbers are shown.
Reptile [31]

Step-by-step explanation:

sorry i couldn't solve the last and the last second one

Was it helpful?

comment down

If there is any kind of suspicious mistake

then sorry

7 0
3 years ago
the artist goes to the store to buy brushes and small cans of paint. he pays $94. he buys 8 brushes that cost $5 each. the rest
sertanlavr [38]
Pays $94
brushes = $40
94 - 40 = 54
54 ÷ 6 = 9
each can of paint costs $9
7 0
3 years ago
Read 2 more answers
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