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3 years ago

12

The freezing point of gasoline is about -50 degrees C, and the freezing point of water is 0 degrees C. What is the difference be

tween the freezing points of gasoline and water?

1 answer:

kow [346]3 years ago

3 0

That water has a greater freezing point then gasoline

You might be interested in

Are these ratios equivalent? 12 friends : 15 strangers AND 6 friends : 14 strangers

meriva

Answer:

No

Step-by-step explanation:

12:15 can be simplified to 4:5

6:14 can be simplified to 3:7

These two are not equivalent to eachother

6 0

3 years ago

Question 1: Four expressions are shown below:

vekshin1

Answers;

Question 1 answer: The first and last option.

Question 1 explanation: 2(4x + 2) is 4x = 12 = 2 = 14 x 2 = 28 and 8x + 4 is 8 x 4 = 32 + 4 = 36 which is 8 more than 28 then, 3x = 9 + 2 + 3 x 2 = 28.

2(4x + 2) = 28 and 8x + 4 equals 36 which is 8 more and they're both equivalent to 2(3x + 2 + x) because 2(3x = 2 = x) equals 28.

6 0

2 years ago

= Knowledge Check

lbvjy [14]

Answer:

Slope is -3.

Step-by-step explanation:

The equation to get the slope is y2-y1/x2-x1. So, plug in the points to the equation. -7-8/7-3 = -15/5 = -3/1 = -3.

3 0

2 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

3 years ago

True or false. - 0.25 = - 1/4

Katena32 [7]

Answer:

true

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers