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dezoksy [38]
3 years ago
6

Write an algebraic expression for the following phrases.

Mathematics
1 answer:
Xelga [282]3 years ago
3 0

Answer:

1. x + 34

2. 2x - 12

3. x - 16

4. x + 45

5. 23 - x

6. 6(x - 7)

7. 78 - 4x

8. 4x + 10

9. 3x + 2

10. x(13 - x)

Step-by-step explanation:

For problems like this in the future, pay attention to the order of words, and how they go together! For example- "the difference of 23 and a number" means to subtract a number from 23, whereas "the difference of a number and 23" would mean to subtract 23 from the number. Words like "increase" and "sum" both mean addition; "product" and "times" both mean multiplication; "difference" and "less" mean subtraction. Just keep these in mind (and watch for context)!

I hope this helped! :)

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4 years ago
Find the domain and range please!
kirza4 [7]

Answer:

domain: -2 < x < 4

range: -∞ < y < 9

Step-by-step explanation:

domain: because the domain is all of the x-values, just look for the furthest one to the left, which is -2, and then look for the one that is furthest to the right, it being 4.

range: because the quadratic function has the two arrows continuing downwards, toward the negative portion of the graph, it would be -∞. Then look for the maximum of the graph, which would be considered the most highest part of the graph, which is seen on the graph as 9.

7 0
3 years ago
Amy paid ?$69.77 for a pair of running shoes during a 25 ?%-off sale. What was the regular? price?
puteri [66]

Answer:

The regular price was $93.03

Step-by-step explanation:

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5 0
4 years ago
Solve: 12^x2+5x-4 =12^2x+6
lyudmila [28]

Answer:

x=2, x=-5

Step-by-step explanation:

to work more comfortably, the first thing we need to do is work the equation linearly, for that we take advantage of the property of logarithms that tells me that log(a^{b})=b*log(a)

in this way, the equation remains as:

log(12^{x^{2} +5x-4 } )=log(12^{5x+6} ) \\ (x^{2} +5x-4 )*log(12)=(5x+6) *log(12)\\x^{2} +5x-4 = 5x+6

Now we clear the equation so that it is of the form ax^{2} +bx+c=0

x^{2} +5x-2x=6+4\\ x^{2} +3x-10=0

finally, we apply the equation to solve second degree equations

x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}

x = \frac{-3 \pm \sqrt {3^2-4*1*(-10)}}{2*1}\\ x = \frac{-3 \pm \sqrt {9+40}}{2}\\ x = \frac{-3 \pm 7}{2}

x=2 and x=-5

Done

5 0
4 years ago
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