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3 years ago

What is the pattern you see when counting by tens

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

8 0

Each number is increased by ten... umm idk any other answer you want

nirvana33 [79]3 years ago

7 0

Only the number in the tens place increases by one and the ones place stays as 0 i.e. 10, 20, 30... you could do 1,2,3... and then just add a zero to the end of that number. when you reach ten keep going, i.e. 8,9,10,11,12,13... would end up being 80,90,100,110,120,130.

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8.97 in word form and expanded form

avanturin [10]

8+.90+.07
This the answer because you have to make it as an addition way.
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3 years ago

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Graph 7/4 on the number line

MakcuM [25]

Answer:

Step-by-step explanation:

First you have to change your fraction:

Change 7/4 into a mixed number

7/4 = 1 3/4

Then you 'guestamate' about 3/4 of the distance between 1 and 2 and mark your point. Like this:

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3 years ago

Will mark as BRAINLIEST

Anon25 [30]

Answer:

B and E

Step-by-step explanation:

B. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 1 do not form right angles.

E. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 4 do not form right angles.

Step-by-step explanation:

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7 0

3 years ago

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A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha

Mariulka [41]

Answer:

V = 63π / 200 m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

$S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx$

- The derivative of the function f'(x) is as follows:

$f'(x) = \frac{21-x}{\sqrt{42x-x^2} }$

- The square of derivative of f(x) is:

$f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }$

- Now use the surface area formula:

$S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi$

- The Volume of the pain coating is:

V = S*t

V = 210*π*3/2000

V = 63π / 200 m^3

8 0

3 years ago

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Ann Landers – who wrote a daily advice column appearing in newspapers across the country – once asked her readers, "If you had i

elena55 [62]

Answer:

It shows that the survey is meaningless because of voluntary response bias.

Step-by-step explanation:

This survey in the question provides a very good example of voluntary response bias and this often overrepresents negative opinions. Now, the people who chose to respond were most likely parents who were very unhappy, and so, there is very little chance that the 10,000 respondents were representative of the population. Knowing more about her readers, or taking a sample of the sample, would not have helped.

Thus, we can conclude that ;The survey is meaningless because of voluntary response bias.

8 0

3 years ago