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AleksandrR [38]
3 years ago
6

Help me Please!! Ive got a deadline!! Thanks! (:

Mathematics
1 answer:
saul85 [17]3 years ago
3 0
The equation in standard form is 2x^2 + 7x - 15=0. Factoring it gives you (2x-3)(x+5)= 0. That's the first one.  The second one requires you to now your formula for the axis of symmetry which is x = -b/2a with a and b coming from your quadratic.  Your a is -1 and your b is -2, so your axis of symmetry is
x= -(-2)/2(-1) which is x = 2/-2 which is x = -1.  That -1 is the x coordinate of the vertex.  You could plug that back into the equation and solve it for y, which is the easier way, or you could complete the square on the quadratic...let's plug in x to find y.  -(-1)^2 - 2(-1)-1 = 0.  So the vertex is (-1, 0).  That's the first choice given.  For the last one, since it is a negative quadratic it will be a mountain instead of a cup, meaning it doesn't open upwards, it opens downwards.  Those quadratics will ALWAYS have a max value as opposed to a min value which occurs with an upwards opening parabola.  This one is also the first choice because of the way the equation is written.  There is no side to side movement (the lack of parenthesis tells us that) so the x coordinate for the vertex is 0.  The -1 tells us that it has moved down from the origin 1 unit; hence the y coordinate is -1.  The vertex is a max at (0, -1)
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2 years ago
you have a liter bottle of orange juice you want to divide the juice into one cup amount how many cups can you pour
Andru [333]
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7 0
3 years ago
Can someone confirm this? is it correct?
EastWind [94]
<span>60 Sorry, but the value of 150 you entered is incorrect. So let's find the correct value. The first thing to do is determine how large the Jefferson High School parking lot was originally. You could do that by adding up the area of 3 regions. They would be a 75x300 ft rectangle, a 75x165 ft rectangle, and a 75x75 ft square. But I'm lazy and another way to calculate that area is take the area of the (300+75)x(165+75) ft square (the sum of the old parking lot plus the area covered by the school) and subtract 300x165 (the area of the school). So (300+75)x(165+75) - 300x165 = 375x240 - 300x165 = 90000 - 49500 = 40500 So the old parking lot covers 40500 square feet. Since we want to double the area, the area that we'll get from the expansion will also be 40500 square feet. So let's setup an equation for that: (375+x)(240+x)-90000 = 40500 The values of 375, 240, and 90000 were gotten from the length and width of the old area covered and one of the intermediate results we calculated when we figured out the area of the old parking lot. Let's expand the equation: (375+x)(240+x)-90000 = 40500 x^2 + 375x + 240x + 90000 - 90000 = 40500 x^2 + 615x = 40500 x^2 + 615x - 40500 = 0 Now we have a normal quadratic equation. Let's use the quadratic formula to find its roots. They are: -675 and 60. Obviously they didn't shrink the area by 675 feet in both dimensions, so we can toss that root out. And the value of 60 makes sense. So the old parking lot was expanded by 60 feet in both dimensions.</span>
8 0
3 years ago
Can sombody please help me with this math problem?
Molodets [167]

y=2/3y-3


1. y = 2/3 (9)-3   multiply 9 and 2 =18/3

 y=18/3-3 divide 18 and 3 =6

y=6-3 subt. 3

y=3



2.  -11=2/3x - 3

 -11+3 = 2/3 x   add 3 to both sides to cancel -3

 -8=2/3 x  

(3/2)-8 = (2/3)3/2 x multiply both sides by reciprical of 2/3

-24/2 = x  divide

-12=x

4 0
3 years ago
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