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tigry1 [53]
4 years ago
14

What is the answer?????

Mathematics
1 answer:
Wewaii [24]4 years ago
8 0
61mm (b)
as the sum of two sides of a triangle is always greater than third
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Find all of the points on the x-axis which are 6 units from the point (−1, 1).
anyanavicka [17]
All the points that are 6 units from (-1, 1) are those on the circle
  (x+1)^2 +(y-1)^2 = 36

For y=0, the two points of interest satisfy
  (x+1)^2 +1 = 36
  (x+1)^2 = 35 . . . . . . subtract 1
  x+1 = ±√35
  x = -1±√35

The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).

5 0
3 years ago
Create and solve a 2 step with subtraction and multiplication Also, x = 9
juin [17]

Answer:

x - 4(x - 2) = -37

x - 4x + 8 = -37

-5x = -37 - 8

-5x = -45

x = 9

4 0
3 years ago
Help with this measurement question pls
Arisa [49]
Around 10.63 because 400/37.61=10.63
3 0
3 years ago
Read 2 more answers
Given that f(x) = 2x −5, find the value of x that makes f(x) = 15. (5 points)
vladimir2022 [97]
First you must set f (x) =15:
15 = 2x - 5
Now you must solve like a normal equation:
15 = 2x - 5 )+5
20 = 2*x ):2
10 = x

ANSWER: The value of x is 10
6 0
3 years ago
Read 2 more answers
Please solve this<br> (Rational numbers 8th grade)
pochemuha

                                        Question # 1

Answer:

\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}

Step-by-step explanation:

Given the expression

\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}

=-\frac{2}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}           ∵  \frac{-2}{3}\times \frac{3}{5}=-\frac{2}{5}

=-\frac{2}{5}+\frac{1}{4}                        ∵   \frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=\frac{1}{4}

\mathrm{Least\:Common\:Multiplier\:of\:}5,\:4:\quad 20

=-\frac{8}{20}+\frac{5}{20}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

=\frac{-8+5}{20}

\mathrm{Add/Subtract\:the\:numbers:}\:-8+5=-3

=\frac{-3}{20}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

=-\frac{3}{20}

Therefore,

\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}

                                               Question # 2

Answer:

\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}

Step-by-step explanation:

Given

\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}

=-\frac{6}{35}-\frac{1}{6}\times \frac{3}{2}\times \frac{2}{5}\times \frac{1}{14}          ∵    \frac{2}{5}\times \frac{-3}{7}=-\frac{6}{35}

=-\frac{6}{35}-\frac{1}{140}         ∵   \frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=\frac{1}{140}

\mathrm{Least\:Common\:Multiplier\:of\:}35,\:140:\quad 140

=-\frac{24}{140}-\frac{1}{140}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

=\frac{-24-1}{140}

\mathrm{Subtract\:the\:numbers:}\:-24-1=-25

=\frac{-25}{140}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

=-\frac{25}{140}

\mathrm{Cancel\:the\:common\:factor:}\:5

=-\frac{5}{28}

Therefore,

\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}

4 0
3 years ago
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