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Alex17521 [72]
3 years ago
8

PLEASEE Can someone help with this? Look at picture

Mathematics
1 answer:
sashaice [31]3 years ago
6 0

Answer:


Step-by-step explanation:

Answer is 72 cm2.For step by step explanation please find the attachment.

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The 7th grade class participated in the following
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20 students

Step-by-step explanation:

Weight lifting:

8% = 0.08

250(0.08) = 20

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Which expression is a difference of squares?
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The answer is
9 {m}^{4}  - 49 {n}^{6}
In the choices; the coefficients 21, 32 and 10 are not perfect squares.
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You have the same bowl, with 5 orange, 6 blue, 3 green, 4 red and 7 yellow candies. You take out an orange candy from the bowl.
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Answer:

9% chance

Step-by-step explanation:

hope this helps.

8 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
Lmk if you can slide answer rn
sattari [20]

A. 70in³ i hope this helps!

3 0
3 years ago
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