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Viefleur [7K]
2 years ago
5

Is knowing the coordinates of the vertex of a parabola enough to determine the domain and range?

Mathematics
1 answer:
swat322 years ago
5 0

Answer:

no

Step-by-step explanation:

You need to know if it's open up or down

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Pls help me I will give brainliest
elixir [45]

Answer:

Step-by-step explanation:

a^8/c

8 0
3 years ago
Read 2 more answers
Which of the two triangles has the greatest area?
katrin [286]

Answer:

they have the same area

Step-by-step explanation:

they have the same measure in legs and the same length hypotenuse. One of them is just transformed either by rotation or refection or translation

5 0
2 years ago
Emily is solving the equation 2(x+9)=4(x+7)+2 where did she make an error
Wittaler [7]
Hello!

2(x + 9) = 4(x + 7) + 2

Distribute the 2

2x + 18 = 4(x + 7) + 2

Distribute the 4

2x + 18 = 4x + 28 + 2

Combine like terms

2x + 18 = 4x + 30

Subtract 18 from both sides

2x = 4x + 12

Subtract 4x from both sides

-2x = 12

Divide both sides by -2

x = -6

Where she did something different than me is her mistake

Hope this helps!
6 0
3 years ago
What is the equation 2X plus 5Y equals 10 what is the table
Assoli18 [71]

Answer:ksksksk

Step-by-step explanation:ggg

5 0
3 years ago
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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