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alex41 [277]
3 years ago
7

The total capacity of two large cups and two small cups is 1,500 milliliters. If the total capacity of one large cup and two sma

ll cups is 1,000 milliliters, what is the capacity of a small cup?
A. 375 milliliters
B. 200 milliliters
C. 500 milliliters
D. 250 milliliters
Mathematics
1 answer:
MissTica3 years ago
8 0

Answer:

The correct answer is option D. 250 milliliters

Step-by-step explanation:

It is given that, the  total capacity of two large cups and two small cups is 1,500 milliliters. If the total capacity of one large cup and two small cups is 1,000 milliliters.

<u>To find the capacity of a small cup</u>

Let small cup capacity  be 's' and large cup capacity be 'l'

From the given information we can write,

2l + 2s = 1500

Then l + s = 750

one large cup and two small cups is 1,000 milliliters.

l + 2s = 1000

l + s + s = 1000

(l + s) + s = 1000

750 + s = 1000

s = 1000 - 750 = 250

The correct answer is option D. 250 milliliters

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The circumference of each of the smaller circles is 12 mm. Find the area of the shaded region.​
podryga [215]

Answer:

11.45mm^2

Step-by-step explanation:

Given data

Circumference= 12mm

The expression for the circumference is

C= 2πr

12= 2*3.14*r

12= 6.28r

r= 12/6.28

r= 1.91mm

Hence the area of each small circle is

A= πr^2

A= 3.14*1.91^2

A= 3.14*3.6481

A=11.45mm^2

8 0
3 years ago
Need Help!
gayaneshka [121]
The discounted price for milk is $3.35

7 0
2 years ago
Simplify and Evaluate
zalisa [80]
\dfrac{2^{-1} + 5^{-1}}{10^{-1} - 5^{-2}} =

= \dfrac{\frac{1}{2} + \frac{1}{5}}{\frac{1}{10} - \frac{1}{5^2}}

=\dfrac{\frac{1}{2} + \frac{1}{5}}{\frac{1}{10} - \frac{1}{25}}

=\dfrac{\frac{5}{10} + \frac{2}{10}}{\frac{5}{50} - \frac{2}{50}}

=\dfrac{\frac{7}{10}}{\frac{3}{50}}

= \frac{7}{10} \times \frac{50}{3}

= \frac{7}{1} \times \frac{5}{3}

= \frac{35}{3}


4 0
3 years ago
Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3
Whitepunk [10]

If f(x)=4x-3:

\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4

If f(x)=4x^{-3}:

\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}

\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}

\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}

7 0
3 years ago
You wanna answer it you get SOME POINTS . AND YOU GET TO MARRY THE GIRL OF YOU'R DREAMS ! ! ! IK right . So do it .
uysha [10]

Answer:

The answer to your question is 20 feet

Step-by-step explanation:

Data

A = (-2, 1)

B = (4, 1)

C = (-2, -3)

D = (4, -3)

Process

1.- Calculate the distance from C and D

dCD = \sqrt{(4 + 2)^{2} + (3 - 3)^{2}}

dCD = \sqrt{6^{2}}

dCD = 6

2.- Calculate the distance from A to C

dAC = \sqrt{(2 - 2)^{2} + (-3 - 1)^{2}}

dAC = \sqrt{4^{2}}

dAC = 4

3.- Calculate the perimeter

Perimeter = 2(dCD) + 2(dAC)

-Substitution

Perimeter = 2(6) + 2(4)

-Simplification

Perimeter = 12 + 8

-Result

Perimeter = 20 ft

3 0
3 years ago
Read 2 more answers
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