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3 years ago

-2 1/2 to -5 3/4 what is the distance between the two on a number line

Mathematics

2 answers:

salantis [7]3 years ago

4 0

Answer:

3 1/4

distances are always positive.. "how many steps would you need to take to walk from -2 1/2 to -5 3/4 on a number like ...

-3,-4,-5 would be three steps , but then you would be at -5 1/2. 1/2 is 2/4's

you are 1/4 away from the -5 3/4 so add another 1/4

Step-by-step explanation:

amid [387]3 years ago

3 0

Answer:

3 1/4

Step-by-step explanation:

difference between the 2 = -5 3/4-(-2 1/2) = -5 3/4+2 1/2 = -3 1/4

since distance is +ve, answer is 3 1/4

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-2/5 - 4/5 + -9/10 = ??? Please help, will mark brainliest for answer with an explanation

NeX [460]

Answer:

Step-by-step explanation:

-2/5 - 4/5 + -9/10

= -2/5 - 4/5 - 9/10

= {-2(2) -4(2) - 9)/10

= (-4 - 8 - 9)/10

= -21/10

= -2 1/10

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3 years ago

Solve each equation. I don't know this

MArishka [77]

1. $(2x - 3)(x + 7) = 0$
 $2x^{2} + 14x - 3x - 21 = 0$
 $2x^{2} + 11x - 21 = 0$

 $x = \frac{-11 +/- \sqrt{11^{2} - 4(2)(-21)}}{2(2)}$

 $x = \frac{-11 +/- \sqrt{121 + 168}}{4}$

 $x = \frac{-11 +/- \sqrt{289}}{4}$

 $x = \frac{-11 +/- 17}{4}$

 $x = -2.75 +/- 4.25$
 $x = -2.75 + 4.25$ $x = -2.75 - 4.25$
 $x = 1.5$ $x = -7$
---------------------------------------------------------------------------------------------------------- 2. $8x(2x - 5) = 0$
 $8x(2x) - 8x(5) = 0$
 $16x^{2} - 40x = 0$
 $16x^{2} - 4x + 0 = 0$

 $x = \frac{-(-40) +/- \sqrt{(-40)^{2} - 4(16)(0)}}{2(16)}$

 $x = \frac{40 +/- \sqrt{1600 - 0}}{32}$

 $x = \frac{40 +/- \sqrt{1600}}{32}$

 $x = \frac{40 +/- 40}{32}$

 $x = 1.25 +/- 1.25$
 $x = 1.25 + 1.25$ $x = 1.25 - 1.25$
 $x = 2.5$ $x = 0$
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3. $x^{2} + 3x - 10 = 0$

 $x = \frac{-3 +/- \sqrt{3^{2} - 4(1)(-10)}}{2(1)}$

 $x = \frac{-3 +/- \sqrt{9 + 40}}{2}$

 $x = \frac{-3 +/- \sqrt{49}}{2}$

 $x = \frac{-3 +/- {7}}{2}$

 $x = -1.5 +/- 3.5$
 $x = -1.5 + 3.5$ $x = -1.5 - 3.5$
 $x = 2$ $x = -5$
----------------------------------------------------------------------------------------------------------
4. $x^{2} = 13x - 36$
 $x^{2} - 13x + 36 = 13x - 13x - 36 + 36$
 $x^{2} - 13x + 36 = 0$

 $x = \frac{-(-13) +/- \sqrt{(-13)^{2} - 4(1)(36)}}{2(1)}$


 $x = \frac{13 +/- \sqrt{169 - 144}}{2}$

 $x = \frac{13 +/- \sqrt{25}}{2}$

 $x = \frac{13 +/- 5}{2}$

 $x = 6.5 +/- 2.5$
 $x = 6.5 + 2.5$ $x = 6.5 - 2.5$
 $x = 9$ $x = 4$
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5. $3x^{2} - 7x + 2 = 0$

 $x = \frac{-(-7) +/- \sqrt{(-7)^{2} - 4(1)(2)}}{2(3)}$

 $x = \frac{7 +/- \sqrt{49 - 8}}{6}$

 $x = \frac{7 +/- \sqrt{41}}{6}$

 $x = \frac{7 +/- 6.403124237432849}{6}$

 $x = 1.167+/- 1.06718737290547$
 $x = 1.67 + 1.06718737290547$ $x = 1.167 - 1.06718737290547$
 $x = 2.73718737290547$ $x = 0.60281262709453$
----------------------------------------------------------------------------------------------------------
6. $10x^{2} - 10x + 9 = 5x^{2} + 4x + 1$
 $10x^{2} - 5x^{2} - 10x + 10x + 9 - 1 = 5x^{2} - 5x^{2} + 4x + 10x + 1 - 1$
 $5x^{2} + 8 = 14x$
 $5x^{2} - 14x + 8 = 14x - 14x$
 $5x^{2} - 14x + 8 = 0$

 $x = \frac{-(-14) + \sqrt{(-14)^{2} - 4(5)(8)}}{2(5)}$

 $x = \frac{14 +/- \sqrt{196 - 160}}{10}$

 $x = \frac{14 +/- \sqrt{36}}{10}$ .

 $x = \frac{14 +/- 6}{10}$

 $x = 1.4 +/- 0.6$
 $x = 1.4 + 0.6$ $x = 1.4 - 0.6$
 $x = 2$ $x = 0.8$

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4 years ago

Last week, Ethan finished 3 1⁄5 books, but this week he only finished 1 4⁄5 books. How many more books did he read last week?

kykrilka [37]

Well first you need to convert the mixed numbers into improper fractions like so

3 1/5 would be 16/5 because 3 times 5= 15+1=16
the same with 1 4/5 but you can do this one a bit differently like so- 1=5/5 + 4/5 = 9/5
So now we move on to the subtraction part of the problem. 16/5-9/5= 7/5 because we know 16-9=7
But now you got to convert this improper fraction back into a proper fraction: 7/ 1 2/5 because 5 goes into 7 one time with a remainder of 2 (2/5)
So the answer would be: Ethan read 1 2/5 more books than he did this week.

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3 years ago

What is the best first step to solve the following quadratic equation?

Basile [38]

Answer:

Step-by-step explanation:

6 0

3 years ago

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GenaCL600 [577]

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