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3 years ago

Find the 70th term of the arithmetic sequence 29, 17, 5

Mathematics

1 answer:

ANTONII [103]3 years ago

7 0

Answer:

857

Step-by-step explanation:

70th term = a + d(n-1)

d= 29-17=12

a=29

n-1 = 69

70th term = 29 + 12×69

= 29 + 828

= 857

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How do I write 435 in expanded from

SVEN [57.7K]

Answer : 400 + 30 + 5

Step - by - step explanation : To write something in expanded form means to write the number individually. " To see the math value of individual digits ".

4 --> Since it's in the hundreds place that means the number is 400 !

3 --> Since it's in the tenths place that means the number is 30 !

5 --> Since it's the first digit that means it stays as the number 5 !

Hope this helped ! <33

Take care happy holidays ! Contact me if you need more help on this ! ( If you want )

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2 years ago

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Rudiy27

a Anna drives her sisters at an a average of 60 mph and devive home

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3 years ago

A cube has a side length of 5xcm. What is the volume of the cube?<br>no links please.

finlep [7]

Answer:

125 cm

Step-by-step explanation:

$V=lwh$

all cube has the same length, width, height

5 * 5 * 5 = 125

Hope this helps

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3 years ago

Friday night, a pizza parlor sold

SpyIntel [72]

The total is. Friday 573

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4 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