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Lera25 [3.4K]
3 years ago
5

Anyoneeee??????????

Mathematics
2 answers:
leva [86]3 years ago
7 0

Your answer will be D

As y = -(x+2)(x-1)

Thus the external negative sign will flip the positive 2 into a negative & negative 1 into a positive; Resulting, y=(x-2)(x+1)

Norma-Jean [14]3 years ago
5 0

We can fairly easily solve this by looking at the direction the graph is pointing, and the x-intercepts.

We see that the graph is pointing down, so we can immediately rule out A and C because they both do not have a -

When we look at the x-intercepts we see they are -2 and 1

Only D when -2 or 1 are plugged in will return an answer of 0

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3 years ago
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Find the value of the variable in each expression
bogdanovich [222]

Answer:

they're all 0

Step-by-step explanation:

N/A

6 0
2 years ago
Radius of cylinder = 5 cm
Kisachek [45]

Answer:

C  1055 04 cm

Step-by-step explanation:

We don't need to see the figure, since we know for sure the cone fits into the cylinder (smaller diameter and height).

So, we first need to calculate the volume of the cylinder, which is given by the formula:

VT = π * r² * h

VT = 3.14 * 5² * 16 = 3.14 * 400 = 1,256 cubic cm

Then we calculate the volume of the cone, which is given by:

VC = (π * r² * h)/3

VC = (3.14 * 4² * 12)/3 = (3.14 * 192)/3 = 200.96 cu cm

Then we calculate the void space left inside the cylinder by subtracting the volume of the cone from the volume of the cylinder:

NV = VT - VC = 1,256 - 200.96 = 1,055.04 cu cm

7 0
3 years ago
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Find the area of the shaded sector of the circle. Leave in terms of pi.
qaws [65]

Answer:

a=10.88\pi

Step-by-step explanation:

using the formula below, the central angle is 20° (bc 360-180-160=20)

radius is 14 (28/2=14)

\frac{20}{360} =\frac{a}{\pi r^2} \\ \frac{1}{18} =\frac{a}{\pi 14^2}

\frac{1}{18} =\frac{a}{196\pi}

a=\frac{1*196\pi}{18} \\a=1*10.88\pi\\a=10.88\pi

3 0
3 years ago
Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81
Anton [14]
Using Lagrange multipliers, we have the Lagrangian

L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)

with partial derivatives (set equal to 0)

L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}
L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}
L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}
L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81

Substituting the first three equations into the fourth allows us to solve for \lambda:

x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}

For each possible value of \lambda, we get two corresponding critical points at (\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3).

At these points, respectively, we get a maximum value of f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3 and a minimum value of f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3.
5 0
3 years ago
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