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GrogVix [38]
3 years ago
9

What two numbers multiply to -96 and add to 20

Mathematics
2 answers:
anygoal [31]3 years ago
6 0
xy=-96\\
x+y=20\\\\
xy=-96\\
x=20-y\\\\
(20-y)y=-96\\
20y-y^2=-96\\
y^2-20y-96=0\\
y^2-24y+4y-96=0\\
y(y-24)+4(y-24)=0\\
(y+4)(y-24)=0\\
y=-4 \vee y=24\\\\
x=20-(-4) \vee x=20-24\\
x=24 \vee x=-4\\\\
(x,y)\in\{(24,-4),(-4,24)\}

So these numbers are -4 and 24.
Reika [66]3 years ago
5 0
If they multiply to -96, that means one number will be negative. If the smaller number of the two is negative, then when you add them you will get a positive number.

Lets first find factors of -96 (just make the smaller number negative so that way when we add we can find the correct sum faster)

-96 = 

96 x -1 
48 x -2 
32 x -3
24 x -4
16 x -6
12 x -8

Now
96 + (-1) = 95
48 + (-2) = 46
32 + (-3) = 29
24 + (-4) = 20
16 + (-6) = 10
12 + (-8) = 4

The only pair that multiplies to -96 and adds to 20 is 24 and -4
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VikaD [51]
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3 years ago
Please help me on the picture above. I will give points for people who help.
Jobisdone [24]

Answer:

I believe the one you selected is correct. A negative plus a negative makes a positive meaning you start at -2 and end with +2

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3 years ago
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HELP PLEASE WILL GIVE BRAINLIST, I DONT HAVE MUCH TIME.IF I PASS THIS I GET TO GO TO DISNEY
Verdich [7]

Answer: 41

Step-by-step explanation:

I think it’s forty one because each child painted one and there are forty one flower pots. Besides that much information wasn’t given so it is most likely 41 sorry if it wasn’t much help

6 0
3 years ago
Which transformations take f(x)=3x+1 to g(x)=−3x-5 ?
lbvjy [14]
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8 0
3 years ago
Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y
DiKsa [7]

The area of the surface is given exactly by the integral,

\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx

We have

y(x)=\dfrac15x^5\implies y'(x)=x^4

so the area is

\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the i-th subinterval are, respectively,

\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2

r_i=\dfrac{5-0}{10}i=\dfrac i2

with midpoint

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

with 1\le i\le10.

Over each subinterval, we interpolate f(x)=\sqrt{1+x^8} with the quadratic polynomial,

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

Then

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that the latter integral reduces significantly to

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)

which is about 651.918, so that the area is approximately 651.918\pi\approx\boxed{2048}.

Compare this to actual value of the integral, which is closer to 1967.

4 0
3 years ago
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