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Taya2010 [7]
3 years ago
5

1. If you are given a fraction, what can you do to create an equivalent fraction? Tell why this method works.

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
6 0
Multiply by 100, and divide by the denominator, then divide by 100 to get a decimal, this works because of a percentage ratio, which is 1.00 to 100.
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Find the slope of line passing through the two points, (7, 1) and (-2, 1).
Mamont248 [21]

Answer: m = 0

Step-by-step explanation:

Slope (m) = y2 - y1/,x2 - x1 : meaning increase in y divided by increase in x . it can also be written as ∆y/∆x

y1= 1, y2 = 1, x1 = 7, x2 = -2

Substitute for those values in the equation above

m = 1 -1/-2 - 7

= 0/-9

Therefore,

m = 0

The slope of the line passing through those coordinates = 0

3 0
4 years ago
Help asap pls and ty
BabaBlast [244]
The answer I got is (-6,-2) the x value is -6
7 0
3 years ago
Read 2 more answers
Evaluate 3^2 + (6-2) × 4 - 6/3​
IrinaVladis [17]
Answer: 23
Explanation: simplify using PEMDAS so solve parentheses, then solve the exponent, multiply the second number by 4, and then add, then subtract.

Hope this helps!
4 0
3 years ago
Simplify the expression and combine like terms. 3(x-2)-2(x+1)3(x−2)−2(x+1)
Nady [450]

Answer:

6x^2+7x+4

Step-by-step explanation:

Explanation is in photo

6 0
2 years ago
X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square
swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = -\frac{4}{5}x+\frac{58}{5}

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

   [x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

   (x + 2)² + (y - 5)²- 29 + 20 = 30

   (x + 2)² + (y - 5)²- 9 = 30

   (x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

    Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

    Slope of the line joining this point to the center (-2, 5),

    m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

          = \frac{10-5}{2+2}

          = \frac{5}{4}

    Let the slope of the tangent which is perpendicular to this line is 'm_{2}'

    Then by the property of perpendicular lines,

          m_{1}\times m_{2}=-1

          \frac{5}{4}\times m_{2}=-1

                 m_{2}=-\frac{4}{5}

   Now the equation of the line passing though (2, 10) having slope m_{2}=-\frac{4}{5}

           y - y' = m_{2}(x-x')

           y - 10 = -\frac{4}{5}(x-2)

           y - 10 = -\frac{4}{5}x+\frac{8}{5}

                  y = -\frac{4}{5}x+\frac{8}{5}+10

                  y = -\frac{4}{5}x+\frac{58}{5}

Therefore, equation of the line will be, y = -\frac{4}{5}x+\frac{58}{5}

7 0
3 years ago
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