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expeople1 [14]
2 years ago
14

(b) In a lab, a substance was heated by 6 °C each hour for 24 hours. What was the total change

Mathematics
1 answer:
WINSTONCH [101]2 years ago
3 0

Answer:

144

Step-by-step explanation:

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PLEASE HELP I HATE THIS SM I WANNA KMSS
dedylja [7]

Answer:

you want to flip the original coordinates and make them both negative so (x,y) would become (-y,-x) so in your problem here Q' would be (0,-7), R' would be (-5,-8), S' would be (-10,-7) and T' would be (-5,-2). Hope that helped :>

Step-by-step explanation:

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3 years ago
Can I please get help with this question please?
solmaris [256]

what are the choices?


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3 years ago
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The formula to determine energy is uppercase E = one-half m v squared. What is the formula solved for v?
alexandr402 [8]

Answer:

Step-by-step explanation:

E=\frac{1}{2}mv^2

All the variables on the right are being multiplied together then the whole mess is being divided by 2.  Let's get rid of the 2 first.  The undoing of division is multiplication, so we will begin by multiplying both sides by 2 to get

2E=mv^2

Next we will move the m. The undoing of multiplication is division.  So we divide both sides by m to get

\frac{2E}{m}=v^2

The undoing of a square is to take the square root, so we will do that to both sides giving us, finally

\sqrt{\frac{2E}{m} }=v

6 0
3 years ago
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Natalie added 20 pieces of candy to a bowl. Two days later, half of the candy in the bowl is gone. If there are 14 pieces left i
Salsk061 [2.6K]

Answer:

8 pieces of candy

Step-by-step explanation:

Take the original amount of candy in the bowl to be x.

It says that 20 pieces of candy were added, therefore there are x+20 pieces of candy in the bowl.

Half of the candy is gone, therefore \frac{x+20}{2} pieces of candy are left in the bowl.

After half of the candy is gone, there are 14 pieces left, so we can say that \frac{x+20}{2} =14

Solve for x:

x+20 = 28

x=8

8 0
3 years ago
Please anyone answer me
ollegr [7]

Let's divide the shaded region into two areas:

area 1: x = 0 ---> x = 2

ares 2: x = 2 ---> x = 4

In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.

Area 1:

A\frac{}{1}  = ∫g(x)dx =  ∫xdx =  \frac{1}{2}  {x}^{2}  | \frac{2}{0}  = 2

Area 2:

A\frac{}{2}  = ∫(g(x) - f(x))dx

= ∫(x -  {(x - 2)}^{2} )dx

=  ∫( - {x}^{2}  + 5x - 4)dx

= ( - \frac{1}{3}{x}^{3} +   \frac{5}{2} {x}^{2}  - 4x)   | \frac{4}{2}

= 2.67 - ( - 0.67) = 3.34

Therefore, the area of the shaded portion of the graph is

A = A1 + A2 = 5.34

3 0
3 years ago
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