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Gemiola [76]
3 years ago
9

At noon, the temperature of a container of ether was –109⁰C. At 1 p.m., the temperature of the container of ether was –34⁰C. Wha

t was the change in temperature from noon to 1 p.m.? 
Mathematics
1 answer:
dem82 [27]3 years ago
5 0
143 degrees celcius because you do 109--34 
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What is the slope of the line represented by the equation y<br> 4 X - 3?<br> 0.-<br> to
Leya [2.2K]

Answer:

The slope is 4/1

Step-by-step explanation:

for every 4 units you go up on the y-axis, you go 1 unit on the x-axis.

6 0
3 years ago
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Hoochie [10]

Answer:

12 hours

hope its right and it helps

Step-by-step explanation:

300/25

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3 years ago
Trevor Has a Fair coin and a six-sided number cube with a number 1 through 6 on each side. He flips the coin and rolls a number
mina [271]

Answer:

1/12

Step-by-step explanation:

When you are doing probability with multiple things at once you have to multiply the probabilities of each asked number from each object together to get the probability of them both landing on those numbers

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3 years ago
The area of the rectangle as a single expression
Ne4ueva [31]

Answer: A=15x^2+6x-48

Step-by-step explanation:

The area of a rectangle can be calculated with this formula:

A=l*w

Where "l" is the length and "w" is the width.

You can identify from the figure that the length and the width of this rectangle are:

length=3(x+2)\\width=5x-8

Then, you need to substitute this lenght and this width into A=l*w:

A=3(x+2)(5x-8)

Now, apply Distributive property:

A=(3x+6)(5x-8)\\\\A=15x^2-24x+30x-48

Finally, you need to add the like terms. Then, you get:

A=15x^2+6x-48

5 0
3 years ago
1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves
erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

\ln|y| = x - \ln|x+1| + C

When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

(b) The curves y = x² and y = 2x - x² intersect for

x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

and the bounded region is the set

\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

7 0
2 years ago
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