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Nadya [2.5K]
3 years ago
5

Explain in a minimum of 2 sentences how to graph the equation of the absolute value function given a vertex of (-1,3) and a valu

e of “a” equal to ½.
Mathematics
1 answer:
liraira [26]3 years ago
7 0
First, we put a point at the vertex. Then, we extend a line upwards of slope 1/2 to the right of the vertex. Then extend a line upwards of slope -1/2 to the left of the vertex. This will give the graph of the equation.
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Alisa had 1/2 L of juice in a bottle. She drank 3/8 L of her juice. What
Elina [12.6K]

Answer:

3/16

Step-by-step explanation:

By drinking 3/8 liters of 1/2 liter juice it means the question tests the ability to multiply fractions. Since she drank 3/8*1/2=3/16

We multiply the numerators and denominators separately ie 3*1/(8*2)=3/16

Therefore, Alisa consumed 3/16 or 0.1875 or 18.75% of the juice.

Since the question was in fractional form, it is prudent to also express answer in fractions hence the right answer is 3/16

3 0
3 years ago
I need to solve cy−7=5d+3y for y, any help?
iVinArrow [24]
Given cy - 7 = 5d + 3y, we solve for y as follows:

cy - 3y = 5d + 7

y(c - 3) = 5d + 7

y = (5d + 7) / (c - 3)
8 0
3 years ago
Read 2 more answers
You need a quart of water but you only have a measuring cup. How many times do you need to fill the cup in order to get a quart?
larisa86 [58]
You need to fill the cup up 4 times since there are 4 cups in a quart. 
6 0
3 years ago
Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis
Tasya [4]
Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5


A.
\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx

B.
sepearte the integrals
\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}

next one
\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}

the last one you can do yourself, it is \frac{50}{3}
the sum is \frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}


so the area under the curve is \frac{233}{3}
6 0
3 years ago
Jeremiah built a box that was 5 inches long by 6 inches wide by 7 inches high .he started filling it up with 1 cube.he realized
Degger [83]
To determine the number of cubes he needs to fill the box (this is assuming the cubes are 1 in cubes, he would need to calculate the volume of the box. To find the volume he would multiply the length by the width by the height. This would be 5 in x 6 in x 7 in. The volume is 210 cubic inches, so he could fill it with 210 one inch cubes.
3 0
3 years ago
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