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I am Lyosha [343]
4 years ago
7

Solve and graph: -2(x-5)>(3x-2)

Mathematics
1 answer:
NARA [144]4 years ago
5 0
     -2(x - 5) > 3x - 2
-2(x) + 2(5) > 3x - 2
   -2x + 10 > 3x - 2
<u> + 2x         + 2x       </u>
             10 > 5x - 2
            <u>+ 2        + 2</u>
             <u>12</u> > <u>5x</u>
              5      5
           2²/₅ > x
              x < 2²/₅
The solution set is equal to (-∞, 2²/₅).
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Please answer both questions and show all work with explanations :) thank you soo much...
torisob [31]

Answer:

1) 109 cm

2) 120 in²

Step-by-step explanation:

A rectangle is a quadrilateral with opposite sides the same length and four (4) right angles. The diagonal, together with two adjacent sides, forms a right triangle. The relationship between side lengths and the diagonal length is given by the Pythagorean theorem. If sides are "a" and "b" and the diagonal is "c", that theorem tells you ...

... c² = a² + b²

1) c² = (60 cm)² + (91 cm)² = 11881 cm²

... c = √11881 cm = 109 cm

2) The perimeter is double the sum of adjacent side lengths, so ...

... 2(a+b) = 46 in

... a + b = 23 in . . . . . divide by 2

The Pythagorean theorem tells you

... a² + b² = (17 in)²

Squaring the equation from the perimeter relation gives ...

... (a + b)² = (23 in)² = a² + 2ab + b²

Subtracting a² + b², we have

... (a² +2ab +b²) - (a² + b²) = (23 in)² -(17 in)²

... 2ab = (529 -289) in² = 240 in² . . . . . simplify

... 240 in²/2 = ab = 120 in² . . . . . area is the product of length and width

_____

We could solve the two equations for a and b to find that there are two possible solutions: (a, b) = (8, 15) or (15, 8). Either way, ab = 8·15 = 120.

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3 years ago
What is the volume of a 575 gram bar of pure gold
ioda
19.32=density D=m÷v so, v=m÷d
v=575÷19.32 v=about 29.8
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4 years ago
Find the limit, if it exists. (if an answer does not exist, enter dne.) lim (x, y) → (0, 0) x4 − 20y2 x2 + 10y2
Firdavs [7]
Traveling along the x-axis, we have

\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-20y^2}{x^2+10y^2}=\lim_{x\to0}\frac{x^4}{x^2}=\lim_{x\to0}x^2=0

On the other hand, along the y-axis we get

\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-20y^2}{x^2+10y^2}=\lim_{y\to0}\frac{-20y^2}{10y^2}=\lim_{y\to0}(-2)=-2

Therefore the limit doesn't exist.
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Answer:

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