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Diano4ka-milaya [45]
3 years ago
13

Freddy poured 1.4 liters of soda pop from a full 2-liter container. How much soda pop was left in the container?

Mathematics
2 answers:
eduard3 years ago
7 0
0.6 liters because 2-1.4 is 0.6
german3 years ago
6 0

Answer:

0.6 liters left

Step-by-step explanation:

So we know that the container is filled with 2 liters.

We also know that he poured 1.4 liters out of the container.

This leads to the conclusion

2 - 1.4 = 0.6

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ASAP!!!!!!!!!!!!!!!!! ANSWER KNOW!!!!!!!!!!!!!!!
koban [17]

Answer:

C

Step-by-step explanation:

C) 3:1 and 9:3

Because when you simplify 9/3, it becomes 3/1 Hence, 3:1

Second question:

1/32 times 60 min = 1.875 min

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Third question:

b) 1152 grams

Hope this helps!

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5 0
4 years ago
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Simplify 8a+3b+8a-4b
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8a+3b+8a-4b=
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7 0
3 years ago
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If you can solve all parts I will give brainliest (also give strategy)
Alexxx [7]

The Halloween conical hat, with given height, circular base and brim

extension has the following calculated parameters;

Part a. The slant height is <u>18.2 inches</u>

Part b. The volume of the cone is 37\frac{1}{2}  \cdot \pi in.³

Part c. The area of the brim, <em>A</em> = <u>36·π in.²</u>

Part d. The area of the brim is found by <u>subtracting the area of the base of the cone from the area covered by the perimeter of the brim</u>

Reasons:

Known parameters;

Height of the conical portion, h = 18 inches

Base circumference, C = 5·π inches

Part a. Slant height of the conical portion; Required

Solution:

The circumference of a circle, C = 2·π·r

Therefore;

r = \dfrac{C}{2 \cdot \pi}

Which gives;

r = \dfrac{5 \cdot \pi}{2 \cdot \pi} = \dfrac{5}{2} = 2.5

Radius, r = 2.5 inches

According to Pythagoras's theorem, we have; s² = r² + h²

Where;

s = The slant height of the cone

s² = 2.5² + 18² = 330.25

s = √(330.25) ≈ 18.2

  • The slant height, <em>s</em> ≈ <u>18.2 inches</u>

Part b. The measure in cubic inches of candy that exactly fills the conical portion of the hat is the volume of the cone.

Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h

Therefore;

V = \dfrac{1}{3} \times \pi \times 2.5^2 \times  18 = 37\frac{1}{2}  \cdot \pi

  • The volume of the cone, V = 37\frac{1}{2}·π in.³

Part c. The extension of the brim from the base of the cone = 4 inches

The radius of the brim, R = Radius of the base of the cone + 4 inches

∴ <em>R</em> = 2.5 inches + 4 inches = 6.5 inches

Area of the brim, <em>A</em> = Area of the 6.5 inch circle - Area of the circular base of the cone

∴ A = π × 6.5² - π × 2.5² = 36·π

  • The area of the brim, <em>A</em> = <u>36·π in.²</u>

Part d. The procedure for solving the question in part c, is described as follows;

  • The area of the brim can be found by finding the entire area of the circle formed by the perimeter of the brim, then subtracting the area of the base of the cone from that area.

Learn more here:

brainly.com/question/17023854

4 0
3 years ago
Which of the following expressions is equivalent to 13^(2)-3^(2)
Amiraneli [1.4K]

Answer:

160

Step-by-step explanation:

13² - 3² = 13 × 13 - 3 × 3 = 169 - 9 = 160

8 0
2 years ago
If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num
weeeeeb [17]

Answer:

<h2>x = 0, y = 5, z = 3</h2>

Step-by-step explanation:

x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right

\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right

\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}

7 0
3 years ago
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