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Gemiola [76]
3 years ago
7

Which equations are true for x = –2 and x = 2? Check all that apply.

Mathematics
1 answer:
4vir4ik [10]3 years ago
4 0
А. х²-4=0; х²=4; х=±2, true. B. x²=-4; there is no solution. C. 3x²+12=0; 3x²=-12; x²=-4; there is no solution. D. 4x²=16; x²=4; x=±2, true. E. 2(x-2)²=0; 2(x²-4x+4)=0; 2x²-8x+8=0; x²-4x+4=0; x1=2; x2=2. False. The answer: A, D.
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Find the midpoint of PQ<br><br> P(4 1/3, 3 1/6), Q(-2 1/5, 3 2/3)
Paha777 [63]

Answer:

(1\frac{1}{15},3\frac{5}{12})

Step-by-step explanation:

To find the midpoint, you must average the x's then average the y's.

So the x's are 4 \frac{1}{3} and -2\frac{1}{5}.

The y's are 3\frac{1}{6} and 3\frac{2}{3}.

To find an average of a data with 2 elements you must add the two elements then take that sum and divide it by 2.

So this is what we will do with the x's then the y's.

So first x:

\frac{4\frac{1}{3}+-2\frac{1}{5}}{2}

Write as improper fractions:

\frac{\frac{13}{3}+\frac{-11}{5}}{2}

Division by 2 is the same as multiplying by 1/2:

\frac{13}{6}+\frac{-11}{10}

The least common multiple of 6 and 10 is 30.  We will multiply first fraction by 5/5 and 3/3 for the second so we can have the same denominator.

\frac{65}{30}+\frac{-33}{30}

Now that the bottoms are the same we can write as one fraction:

\frac{65+-33}{30}

Simplify:

\frac{32}{30}

If you want the answer as a mix fraction like your question began as we can do that by first figuring out how many 30's are in 32 and what is the remainder of that division.

There is one 30 and 32 and so 32-30=2 is the remainder.

1\frac{2}{30}

Reduce by dividing top and bottom by 2:

1\frac{1}{15}

Now for y:

\frac{3\frac{1}{6}+3\frac{2}{3}}{2}

Write the mix fractions as improper fractions:

\frac{\frac{19}{6}+\frac{11}{3}}{2}

Dividing by 2 is the same as multiplying by 1/2:

\frac{19}{12}+\frac{11}{6}

The least common multiple of 12 and 6 is 12 so I'm going to multiply the second fraction by 2/2 so the denominators will be the same:

\frac{19}{12}+\frac{22}{12}

Since the denominators are the same we can write as a single fraction:

\frac{41}{12}

How many 12's are in 41? 3 and so the remainder is 41-3(12)=41-36=5.

3\frac{5}{12}

So the midpoint is:

(1\frac{1}{15},3\frac{5}{12}).

6 0
3 years ago
Read 2 more answers
Jarvis left out a statement from the proof shown.which statement is best explained by the corresponding angles postulate
Salsk061 [2.6K]
The postulate of the corresponding angles establishes that when a transversal line cuts two parallel lines, the corresponding angles are congruent. These angles are on the same side of the parallel lines and on the same side of the transversal line.

  Then, if we based on this definition and analize the figure attached, we can notice that the angles ∠1 and ∠3 are corresponding angles, so they are congruent. In this case the angle ∠1 is internal and the angle ∠3 is external.

 The answer is: ∠1 and ∠3 are congruent (See the image attached).

7 0
3 years ago
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Jack usually mows his lawn in 6 hours. marilyn can mow the same yard in 4 hours. How much time would it take for them to mow the
Anuta_ua [19.1K]

Answer:

2.4 hours or 2 hours and 24 minutes

Step-by-step explanation:

Jack mows 1/6 of the lawn per hour while Marilyn mows 1/4 of the lawn together. Adding up those rates yields in their combined mowing rate:

R = \frac{1}{6}+ \frac{1}{4}\\R=\frac{10}{24}

The time required for them to mow the entire lawn is:

t = \frac{1}{R}\\t = \frac{1*24}{10}\\ t=2.4\ hours

It would take them both 2.4 hours to mow the lawn together.

5 0
3 years ago
Numbers like 10, 100, 1,000, and so on are <br><br> called?
mylen [45]
Numbers starting with a 1 and followed by only 0s (such 10, 100, 1,000,10,000, and so forth) are called powers of ten, and they're easy to represent as "exponents". Powers of ten are the result of multiplying 10 times itself any number of times.
3 0
3 years ago
8.<br> (19x - 18)<br> (10x -9<br> (7x+1)
harkovskaia [24]

Answer:

162

Step-by-step explanation:

assuming you forgot the parentheses at the end of '9'. Do you want an explanation?

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3 years ago
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