Min wants to make 100 name tags with ribbons attached to them. Each name tag requires five centimeters of ribbon. She has 3.25 meters of ribbon. Exactly how many more centimeters of ribbon does Min still need to make 100 name tags?
A: 175
B: 305
C: 325
D: 825

Do it Step By Step

Answer:

ok 12 is 12 1 is 1 2 is 2 and three is treee

Answer:

150 m²

Step-by-step explanation:

Given 1 cm represents 5 m

⟹1 cm=5 m

given area =6 cm² =6×(5 m)²  =6×25 m²

=150 m²

area of 6 cm² represents 150 m²

Answer:

40%

Step-by-step explanation:

divide 80 by 100 80/100= 0.8 then divide 32 by 0.8 to get 40%

Answer:

C. d = C/pi

Step-by-step explanation:

lol nothing really to explain but it's saying diameter equals circumference divided by pi

Hope this helps dude !

Answer:

0.10

Step-by-step explanation:

If one meter is 100 centimeters and you're trying to find how much of a meter is in 10 centimeters all you need to do is divide the meter by 10, just like they did to get the amount of centimeters.

Answer :

⠀

Step-by-step explanation:

⠀

[tex] {\longrightarrow \: \sf \: Let \: \: a : b\: = 3: 4 = \dfrac{a}{b} \: = \: \dfrac{3}{4} = k}[/tex]

⠀

Therefore,

⠀

[tex] \dashrightarrow \qquad \sf \: a = 3k[/tex]

[tex] \dashrightarrow \qquad \sf \: b = 4k[/tex]

⠀

Now, Substituting the values :

⠀

[tex]\longrightarrow \qquad \sf \: 15a+9b:16a+5b[/tex]

⠀

[tex]\longrightarrow \qquad \sf \: \dfrac{15(3k)+9(4k)}{16(3k)+5(4k)}[/tex]

⠀

[tex]\longrightarrow \qquad \sf \: \dfrac{45k+36k}{48k+20k}[/tex]

⠀

[tex]\longrightarrow \qquad \sf \: \dfrac{81k}{68k}[/tex]

⠀

[tex]\longrightarrow \qquad \sf \: \dfrac{81 \cancel{k}}{68 \cancel{k}}[/tex]

⠀

[tex]\longrightarrow \qquad \sf \: \dfrac{81}{68}[/tex]

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[tex]\longrightarrow \qquad \bf \: {81} : {68}[/tex]

⠀

Therefore ,

⠀

[tex]{ \pmb{ \sf{\longrightarrow \qquad \: 15a+9b:16a+5b = 81 : 68}}}[/tex]

⠀

Answer:

-8c + 6

Step-by-step explanation:

2 (-6c + 3) + 4c

Distribute the 2 with the expression in the parenthesis:

-12c + 6 + 4c

*Parenthesis are no longer needed

Combine Like terms:

-8c + 6

*Ask yourself is it in its simplest form? If no, continue your work. If yes, you have your answer.

The total amount of money the server would have saved after 5 weeks is  $455.

The first step is to determine the total amount of money saved for the first four weeks.

Total amount of money saved in the first four weeks = 35% x (200 x 4) = $280

Amount saved on the 5th week = 7/8 x $200 = $175

Total savings = $175 + $280 = $455

To learn more about fractions, please check: https://brainly.com/question/1301963

Answer:

465

Step-by-step explanation:

I just did it

Answer:

Step-by-step explanation:

a total of 360 tickets were sold for the school play. they were either adults tickets or student tickets. the number of student tickets sold was two times the number of adult tickets sold. how many adult tickets were sold?

Answer by chibabykalu1 (Show Source): You can put this solution on YOUR website!

x, =student tickets

y, =adult tickets

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exact translation of the description into a system of equations. Solve for the value of y.

The process can be, solve for x in terms of y, already seen as THE second equation of the system; substitute into the other equation, and solve the result for y, and compute.

Substitute y = 120 into equation 2

Answer:    0.435 repeating

Step-by-step explanation:

Call the amount of chips ordered : x

-> The amount of hot dogs ordered : 4x

Since there are a total of 20 chips and hot dogs orders, x + 4x = 20 -> 5x = 20

-> x = 20/5 = 4
-> 4x = 4 x 4 = 16

So, there are 16 hot dogs and 4 chips ordered.

Answer:

16 hot dogs

4 chips

Step-by-step explanation:

Let h = number of hot dogs ordered

Let c = number of chips ordered

Given:

⇒ h + c = 20

⇒ h = 4c

Substitute  h = 4c  into  h + c = 20  and solve for c:

⇒ 4c + c = 20

⇒ 5c = 20

⇒ c = 4

Substitute found value for c into  h = 4c  and solve for h:

⇒ h = 4 × 4

⇒ h = 16

Therefore, 16 hot dogs and 4 chips were ordered.

Answer:

What color does the letter purple taste like?

Step-by-step explanation:

Changing the subject of an equation involves solving for another variable in the equation

The equation of t is [tex]t=\frac{\sqrt{y_i-y_o}}{N}[/tex]

The equation is given as:

[tex]N=\sqrt[2]{\frac{y_i-y_o}{t^2}}[/tex]

Rewrite the equation as:

[tex]N=\sqrt{\frac{y_i-y_o}{t^2}}[/tex]

Evaluate the square root of t^2

[tex]N=\frac{\sqrt{y_i-y_o}}{t}[/tex]

Multiply both sides by t

[tex]Nt=\sqrt{y_i-y_o}[/tex]

Make t the subject, in the above equation

[tex]t=\frac{\sqrt{y_i-y_o}}{N}[/tex]

Hence, the equation of t is [tex]t=\frac{\sqrt{y_i-y_o}}{N}[/tex]

Read more about subject of formula at:

https://brainly.com/question/657646

Answer:

50.7016

Step-by-step explanation:

1.Area of rhombus is l^2

area of rhombus=1/2*d1*d2

Answer:

the answer is C.25.2

Step-by-step explanation:

i took the quiz and got it right.

Answer:

Step-by-step explanation:

B

A

A

Answer: x = 10

Step-by-step explanation: hope this helps

The volume of the sphere that has a radius of 3 inches is: 113.1 in.³

Volume of a sphere is found using the formula, V = 4/3 πr³, where r is the radius of the sphere.

Given the following:

Plug in the values into V = 4/3 πr³:

Volume of the sphere = 4/3 (3.14)(3)³

Volume of the sphere = 113.1 in.³

Learn more about the volume of a sphere on:

https://brainly.com/question/20394302

Check the picture below, so the parabola looks more or less like that.

now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.

[tex]\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7[/tex]

Answer:

48 i think

Step-by-step explanation:

Answer:

$48

Step-by-step explanation:

[tex]e^{x^2y} - e^y = e^y(e^{x^2} - 1) = x[/tex]

We should ISOLATE x

[tex]e^y= \frac{x}{e^{x^2} - 1}[/tex]

Find the Natural Log of Both Sides to Make the Left Side "y"

[tex]y = ln(\frac{x}{e^{x^2}-1})[/tex]

Now, FIND THE DERIVATIVE Using Chain Rule!!!

[tex]y' = \frac{1}{\frac{x}{e^{x^2}-1}} * \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2}= {\frac{e^{x^2}-1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2} = {\frac{1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)} = {\frac{1}{x}}* \frac{(e^{x^2}-1 - 2x^2e^{x^2})}{(e^{x^2}-1)}[/tex]

Answer:

The kilograms of chocolate for the 40% fat content (1st chocolate) = 90kg

The kilograms of chocolate for the 50% fat content (2nd chocolate) = 10kg

Step-by-step explanation:

A candy distributor needs to mix a 40% fat-content chocolate with a 50% fat-content chocolate to create 100 kilograms of a 41% fat-content chocolate. How many kilograms of each kind of chocolate must they use?

Let us represent

Kilograms of first chocolate = x

Kilograms of second chocolate = y

A candy distributor needs to mix a 40% fat-content chocolate with a 50% fat-content chocolate to create 100 kilograms of a 41% fat-content chocolate.

Hence:

40% × x + 50% × y = 41% × 100

0.40x + 0.50y = 0.41× 100

0.40x + 0.50y = 41.......Equation 1

x + y = 100kg ......Equation 2

x = 100 - y

Substitute 100 - y for x

0.40(100 - y) + 0.50y = 41

40 - 0.40y + 0.50y = 41

Collect like terms

- 0.40y + 0.50y = 41 - 40

0.10y = 1

y = 1/0.1

y = 10.

Therefore, the kilograms of chocolate for the 50% fat content (2nd chocolate) = 10kg

Remember:

x = 100 - y

x = 100kg - 10kg

x = 90kg

The kilograms of chocolate for the 40% fat content (1st chocolate) = 90kg

Answer:

X=9

Hope this helps! :)

The value of h'(1) is -8

Given the following expression;

[tex]h (x) = 5 f(x) - 4g (x) + 3x^2 - 2[/tex]

Differentiating the function above:

[tex]h' (x) = 5 f'(x) - 4g' (x) + 6x[/tex]

Given the following parameters

[tex]f' (1) = 2\\g' (1) = 6[/tex]

Get h'(1)

[tex]h' (1) = 5 f'(1) - 4g' (1) + 6(1)\\h' (1)=5 (2) - 4(6) + 6(1)\\h' (1)= 10-24+6\\h' (1)=-8[/tex]

Hence the value of h'(1) is -8

Learn more here: https://brainly.com/question/1557484

Answer:

5) 0 6 ) 1/4

Step-by-step explanation:

So the slope ( gradient ) of line E is 0 because it is straight and it as no slope

The slope of line F is 1/4 because:

Gradient = y^2 - y^1 / x^2 - x^1

Gradient = 2-0 / 6- -2

Gradient= 2 / 8

2/8 in lowest term is 1/4

YOU MIGHT AVE A DIFFERENT METHOD BUT ANSWERS MUST BE THE SAME.

HOPE IT HELPED

Answer:

43.2 pounds of fish :)

Step-by-step explanation:

120 x 0.36 = 43.2

Have a great day!!

Please rate and mark brainliest!!!

Answer:

7.561 – 2.37=5.191 if you wanted me to show work, comment and ill change it.

Step-by-step explanation: