Mass = 100g, volume = 20cm^3, what is the density?

Answer:

Of (multiplication) 0*-1=0, sum (addition) 0+(-1)=-1

Step-by-step explanation:

Answer:

$8.76*x

Step-by-step explanation:

Answer:

xy=1.5

-7x=1.75

x+2y=-12.25

(1/3)y=-2

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

34.3%

Step-by-step explanation:

Answer: 17600

Step-by-step explanation: 16000*10% is the same as 16000*0.1, then you get 1600. 16000+ 1600 is 17600

Answer:

-1/2

Step-by-step explanation:

Answer:

-1/2

Step-by-step explanation:

convert mixed fraction to impropoer fraction

Answer:

goods and services

money

Step-by-step explanation:

Answer:

The answer is A. Money and D. Goods and Services.

I took the test

Answer:

an addition problem

Step-by-step explanation:

Answer:

estimated 260 seventh graders use computer 5 to 10 hours a week and 300 seventh graders for 10 hours or less a week

Step-by-step explanation:

260 out of 300

300 our of 360

these numbers because it's an estimate it's what you think there is no correct answer just an estimation

Answer:

TWU is congruent to WUV

(I actually don't know if that's how I'm supposed to answer it but do what you will)

Answer:

The volume of the metal used is 5277.9 cm³.

Step-by-step explanation:

The volume of a cylinder is given by:

[tex] V = \pi r^{2} h [/tex]

Where:

r: is the radius

h: is the height = 50 cm

The volume of metal ([tex]V_{m}[/tex]) can be found by subtracting the internal volume ([tex]V_{i}[/tex]) from the external volume ([tex]V_{e}[/tex]):

[tex] V_{m} = V_{e} - V_{i} [/tex]

[tex] V_{m} = \pi r_{e}^{2} h - \pi r_{i}^{2} h [/tex]    

The external radius is given by the sum of the internal radius with the tickness:

[tex] r_{e} = t + r_{i} [/tex]

Hence, the volume is:

[tex] V_{m} = \pi*50 cm((2.4 cm + 5.8 cm)^{2} - (5.8 cm)^{2}) = 5277.9 cm^{3} [/tex]

Therefore, the volume of the metal used is 5277.9 cm³.

I hope it helps you!                                                      

Answer:

[tex]2^{9}[/tex]

Step-by-step explanation:

Using the rule of exponents

[tex](a^m)^{n}[/tex] = [tex]a^{mn}[/tex]

note that 8 = 2³ and so

8³ = (2³ )³ = [tex]2^{3(3)}[/tex] = [tex]2^{9}[/tex]

Answer:

congruent by side angle side so the triangle is congruent

Answer:

Option B 540 is the amswer

Answer:

58

Step-by-step explanation:

first we want to write an equation and that equation would be

2500=40x+160 now we solve for x

step 1 subtract 160 from each side

2340=40x

then we divide each side by 40

x=58.5

because he cant bring half a box he can only bring 58

Answer:

one side = 8xy²·[tex]\sqrt{2x}[/tex]

Step-by-step explanation:

if the area is 128x^3y^4 then to find one side you must take the square root

[tex]\sqrt{128}[/tex][tex]\sqrt{x^3}[/tex][tex]\sqrt{y^4}[/tex] = [tex]\sqrt{4}[/tex][tex]\sqrt{32}[/tex][tex]\sqrt{x^2}[/tex][tex]\sqrt{x}[/tex][tex]\sqrt{y^2}[/tex][tex]\sqrt{y^2}[/tex] =

2xy²[tex]\sqrt{16}[/tex][tex]\sqrt{2}[/tex][tex]\sqrt{x}[/tex] =

8xy²·[tex]\sqrt{2x}[/tex]

Answer:

the Answer is 2 hope that helps

Answer:

yes

Step-by-step explanation:

Answer: .209

Step-by-step explanation:

took the test

Answer:

1. 24.36

2. Do the same thing (follow the steps in the picture)

Step-by-step explanation:

Answer:

3796.65

Step-by-step explanation:

I = $ 3,106.35

Equation:

I = Prt

Calculation:

First, converting R percent to r a decimal

r = R/100 = 3.9%/100 = 0.039 per year,

then, solving our equation

I = 17700 × 0.039 × 4.5 = 3106.35

I = $ 3,106.35

The simple interest accumulated

on a principal of $ 17,700.00

at a rate of 3.9% per year

for 4.5 years is $ 3,106.35.

$ 6,903.00 - 3,106.35 = 3796.65

Answer:

36$

24/2*3=$36

Answer:

$18

Step-by-step explanation:

The game costs $24, and so far he has saved three- fourths of the cost.

3/4(24)=x

A number that would lie between -2.15 and -2.16 on the number line are 2.151.

Integers are arranged at equal intervals along horizontal straight lines called number lines. A number line can be used to represent every number in a series. The endpoints of this line go on forever. Some will have a defined beginning and end. It refers to those as closed number lines. Number lines might be alternatively left empty or blank. A number ladder is a vertical number line.

There are some possibilities would include the following:

2.151, 2.152, 2.153, 2.154, 2.155, 2.156, 2.157, 2.158, 2.159

therefore, the numbers that are between 2.15 and 2.16 are 2.151.

Learn more about number line here:

brainly.com/question/13425491

#SPJ2

Answer:

I think it's would be D

Step-by-step explanation:

1/2 is 50% So you add 19 and 50 to get 69% or 69/100

Answer:

x=8\3................

I don't know.... sorry...!...

Answer: B

Step-by-step explanation:

3x - 4 = 11

add 4 to both sides so it cancels out

3x = 15

Now divide by 3

x = 5

Answer:

x=5

Step-by-step explanation:

3x - 4 = 11

4 + 11 = 15

3x = 15

x = 5

9514 1404 393

Answer:

  6.  A

  7.  C

  8.  E

Step-by-step explanation:

6. Devon gets $0.45 for each of the s programs he sells, so his pay for selling is 0.45s. He gets a $40 flat rate for each game, so his total pay is ...

  p = 0.45s + 40 . . . . . matches choice A

__

7. Robert gets $25 per hour in tips for each of the x hours he works, so his tip earnings are 25x. He gets a $40 flat rate for each day he works, so his total pay for working x hours is ...

  y = 25x +40 . . . . . matches choice C

__

8. The graph of a proportional relationship is a straight line through the origin. This is described by B and C, so choice E is the best.

Answer:

tan(A−B)=tan(A)−tan(B)/1+tan(A)tan(B)

1+tan(A)tan(B) To prove this, we should know that:

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)AND

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)We start with:

1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)We start with:tan(A−B)=sin(A−B)/cos(A−B)

cos(A−B)tan(A−B)=sin(A)/cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.Dividing numerator and denominator in (1) by cos(A)cos(B) we get:

cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.Dividing numerator and denominator in (1) by cos(A)cos(B) we get:tan(A−B)=sin(A)/cos(B)−cos(A)/sin(B)/;2cosA)cos(B)/cos(A)cos(B)+sin(A)sin(B)/cos(A)cos(B)

cos(A)cos(B)tan(A−B)=sin(A)cos(B)/cos(A)cos(B)−cos(A)sin(B)cos/(A)cos(B)cos/(A)cos(B)co(A)cos(B)+sin(A)sin(B)/cos(A)cos(B)

cos(A)cos(B)tan(A−B)=sin(A)/cos(A)−sin(B)cos(B)1+sin(A)cos(A)sin(B)cos(B)

cos(A)−sin(B)cos(B)1+sin(A)cos(A)sin(B)cos(B) tan(A−B)=tan(A)−tan(B)/1+tan(A)tan(B)

Step-by-step explanation:

Hope it is helpful....