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3 years ago

11

pls help me asap and fast and for every question, you have to explain to me how you got the answer the one who did all of the wi

ll get a brainly

1 answer:

shtirl [24]3 years ago

5 0

Answer:

180-12-35

----------

30

= 133

----------

30

3×2×1

----------

2 × 2

=6/4

=3/2

6(2÷3/5)

=6(10/3)

=20

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50× 3/8 how to reduce answer to lowest te

hoa [83]

The lowest term is $\frac{75}{4}$ .

Solution:

Given expression is $50\times\frac{3}{8}$

<u>To reduce this term to the lowest term:</u>

$$50\times\frac{3}{8}=\frac{50}{1}\times\frac{3}{8}$

Multiply the numerator and denominator.

$$50\times\frac{3}{8}=\frac{150}{8}$

Now, divide the numerator and denominator by the greatest common factor.

Here 150 and 8 both have common factor 2.

So, divide numerator and denominator by 2.

$$=\frac{150\div2}{8\div2}$

$$=\frac{75}{4}$

$$50\times\frac{3}{8}=\frac{75}{4}$

Hence the lowest term is $\frac{75}{4}$ .

4 0

3 years ago

Can somebody help me ???

Ann [662]

It is D because if you look at the graph every 5 years the acres decrease by 20

6 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

8n = -3m + 1; n = -2, 2, 4

bogdanovich [222]

I am assuming you want to solve for m in each case

8n = -3m + 1

8(-2) = -3m + 1

-16 = -3m + 1

-3m = -17

m = $\frac{17}{3}$

8(2) = -3m + 1

16 = -3m + 1

-3m = 15

m = -5

8(4) = -3m + 1

32 = -3m + 1

-3m = 31

m = $\frac{31}{3}$

8 0

3 years ago

Read 2 more answers

What is 3 minus 3 times 6 plus 2?

Zina [86]

What is 3 minus 3 times 6 plus 2?

Write it out:-

3 - 5 × 6 + 2 = ?
3 - 30 + 2 = ?
-27 + 2 =
-25

<span>3 minus 3 times 6 plus 2 is -25. </span>

7 0

3 years ago

Read 2 more answers