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3 years ago

Decimals- do i move the decimal when adding or multiplying

Mathematics

1 answer:

EastWind [94]3 years ago

6 0

Answer:

when adding you just line the decimals up like this, let's say you had to add 1 to 2.58.

2.58

+ 1 .00

3.58

when multiplying you might have to move it, say you had to multiply 1.5 by 2.2. (you don't have to line up the decimals like with addition)

1 .5

× 2.2

+ 300

3.30

1.5 has one decimal place and 2.2 also has one decimal place so you put those 2 decimal places in your answer.

sorry if this is a bit confusing lol

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What is the approximate volume of a cone with a height of 9 ft and radius of 3 ft?

tensa zangetsu [6.8K]

Volume of a cone=1/3 times height times area of base
area of base=area of circle=pi times r^2
r=3
aprox pi to 3.14
height=9

so we have
volume=1/3 times 9 times 3.14 times 3^2
3^2=9
1/3 times 9 times 3.14 times 9
3 times 3.14 time s9
9.42 times 9
84.78
nearest tenth
0.78=0.8

answer is 84.8 ft^3

3 0

3 years ago

Read 2 more answers

Mrs.Kennedy is buying pencils for each of 315 students at Hamilton Elementary. The pencils are sold in boxes of tens. How can sh

g100num [7]

Mrs Kennedy is buying pencils for 315 students.

The pencils are sold in boxes of 10.

Therefore the number of pencils she needs to buy should be a multiple of 10, so that she can buy boxes of 10 pencils each.

number of students are 315 but 315 is not a multiple of 10

so we have to round off 315 to the nearest ten, that's 320.

so then she has to buy 320 pencils

number of boxes she needs - 320 / 10 = 32 boxes of pencils

she only needs pencils for 315 students so she will have 5 extra pencils

she will have to buy 32 boxes

4 0

3 years ago

Read 2 more answers

13 POINTS- please help me

allsm [11]

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures $(6x+20)^{\circ}$ and $(x+100)^{\circ}$ are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

$6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16$

Then

$(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}$

17. Two parallel lines are cut by transversal. Angles with measures $(2x+12)^{\circ}$ and $(3x-22)^{\circ}$ are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

$2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34$

Then

$(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}$

18. Two parallel lines are cut by transversal. Angles with measures $(6x-7)^{\circ}$ and $(5x+10)^{\circ}$ are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

$6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17$

Then

$(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}$

19. The diagram shows two complementary angles with measures $2x^{\circ}$ and $56^{\circ}$ . The measures of complementary angles add up to $90^{\circ},$ then

$2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17$

Hence,

$2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}$

Check:

$34^{\circ}+56^{\circ}=90^{\circ}$

20. Angles $\angle 1$ and $\angle 2$ are vertical angles. By vertical angles theorem, vertical angles are congruent, so

$m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4$

Hence,

$m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}$

21. $\angle 5$ and $\angle 8$ are supplementary. The measures of supplementary angles add up to $180^{\circ},$ so

$m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34$

Therefore,

$m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}$