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alina1380 [7]
3 years ago
5

Whats the answer to this?

Mathematics
1 answer:
Anarel [89]3 years ago
8 0

Answer: So the answer is BStraight lines are produced by linear functions. That means that a straight line can be described by an equation that takes the form of the linear equation formula thats one thing i found online

Step-by-step explanation: use this formula y=mx +b/. Here's one more thing that can be useful; The graph below represents any line that can be written in slope intercept form. It has two slider bars that can be manipulated. The bar labeled m lets you adjust the slope, or steepness, of the line. The bar labeled b changes the y-intercept. Try sliding each bar back and forth, and see how that affects the line.

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Two cylinders have equal diameters. are their volumes also equal? is the same true for cones and spheres?
Paraphin [41]
<span>Although two cylinders may have equal diameters, their volumes are not necessarily equal, as the volume of a cylinder is dependent also on the cylinder's height. The same holds true for cones. Spheres, however, are, by nature, proportional. Therefore, if two spheres have the same diameter, they also have the same volumes.</span>
5 0
3 years ago
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(HELP FAST PLEASE) g^2-9 Factor each expression. Show your work
vovangra [49]

Answer:

(g+3)(g-3)

Step-by-step explanation:

square root of g squared is g

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4 years ago
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musickatia [10]

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3 years ago
Find the LCM of 3,12,16​
maria [59]

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3 years ago
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A frog leaps 24 inches. The highest point in the jump is 6 inches. Assume the frog starts at (0,0). What quadratic function mode
Ivanshal [37]
The path is in the shape of a parabola, the horizontal length is 24, so the middle point is at x=12, the symmetry line is x=12, the highest point (the vertex) is at (12,6)
the equation in vertex form is y=a(x-12)²+6
next, find a by using either one of the two points, the starting point (0,0) or the end point (0,24). obviously (0,0) is easier to calculate:
0=a(0-12)² +6
a=-1/24
so the quadratic equation is y=-\frac{1}{24} (x-12)^2+6
5 0
4 years ago
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