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Temka [501]
3 years ago
6

Help me solve this, for 15 points, I need it for my homework teehee.

Mathematics
1 answer:
faust18 [17]3 years ago
5 0

Answer:

its wrong because verticle angles would need to be opposite of each other, not right next to each other. ACB and FCE would be correct

Step-by-step explanation:

im not too sure how to explain it, verticle angles are across from each other

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Find all points having an x-coordinate of 2 whose distance from the point (-1,-2) is 5
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5=\sqrt{(-1-2)^2+(-2-y_1)^2}\\
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y_1^2+4y_1+13=25\\
y_1^2+4y_1-12=0\\
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y_1(y_1+6)-2(y_1+6)=0\\
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3 0
3 years ago
Please help me i need help points and brainly
snow_lady [41]
The values go up by 2. So your answer would be 20.
4 0
3 years ago
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Does anyone know how to turn a fraaction percent into a decimal like this:
lilavasa [31]
3 divided by 8 and 17 divided by 30
4 0
3 years ago
Mr. Evens Is paid $9.20 per hour for the first 40 hours he works in a week he is paid 1.5 times that rate for each hour after th
spin [16.1K]
I don't agree with his statement. The first part says that he was paid $9.20 for the first 40 hours so we multiply them (9.20 x 40) to get 368. Then it says that he was paid 1.5 times that rate for every hour after that. The rate is "$9.20 per hour" so you find 1.5 times $9.20. You multiply $9.20 by 1.5 to get $13.80 per hour. Now you have to find how much money he got after the initial 40 hours. It says he worked 42.25 hours, and we already figured out how much money he got the first 40 hours ($368). So for the remaining 2.25 hours, Mr.Evens is paid $13.80 an hour. So we multiply 2.25 by $13.80 to get $31.05. To find out the total amount of money accumulated, we add. $368 + $31.05 is $399.05 which means that Mr.Evens is not correct. Hope this helped :]
7 0
3 years ago
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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 :

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

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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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