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professor190 [17]
3 years ago
12

Which equation(s) represent joint variations?

Mathematics
2 answers:
Helen [10]3 years ago
8 0

Answer- It's the last two, D & E

Step-by-step explanation:

ale4655 [162]3 years ago
4 0

Answer:

d and e

Step-by-step explanation:

You might be interested in
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
3 years ago
Write the prime factorization of 10. Use exponents when appropriate and order the factors from least to greatest (for example, 2
inna [77]

Answer:

2 x 5

Step-by-step explanation:

in prime factorization, we divide until it gets to 1. In the case of 10 divide by 2 which will give you 5 divide by 5. Thats how I got 2 x 5

6 0
3 years ago
4 out of 12 family riders are for young readers under the age of 5 what fraction has the same value as4/12?
nataly862011 [7]
The answer would be 1/3 because if you divide 4 divided by 4 it's 1 and 12 divided by 4 is 3!
8 0
3 years ago
Read 2 more answers
Please help me plzzzzzz​
erma4kov [3.2K]

Answer:

976 square cm

Step-by-step explanation:

Area of one 2d triangle face: (24*14)/2=168 square cm

Area of both 2d triangles faces: 168 + 168 =336 square cm

Area of one side: 25*10=250 square cm

Area of both sides: 250+250=500

Area of base: 14*10=140 (since it's a triangular prism, there's only one base)

Surface Area: 336+500+140=976 square cm

Hope this helps! :)

3 0
3 years ago
In the figure, point B is the midpoint of AC . Use the figure to answer the questions.
kirza4 [7]

Answer

Proof of (a)

Disagree with the  ΔABD = ΔCBD by the SAS congruence postulate .

SAS congurence property

In this congurence property two sides and one angle of the two triangles are equal .

(1) BD = BD ( Common sides of the  ΔABD and ΔCBD )

(2) AB =BC (B is the midpoint of the AC i.e it bisect AC in the two equal parts.)

but equal angles are not given

therefore disagree with the  ΔABD = ΔCBD  by the SAS congurence postulate .

Proof of (b)

SSS congurence property

In this congurence property three sides of the two triangles are equal .

In the ΔABD and ΔCBD .

(1) BD = BD ( Common sides of the  ΔABD and ΔCBD )

(2) AB =BC (B is the midpoint of the AC i.e it bisect AC in the two equal parts.)

(3) AD=CD  (Given )

ΔABD = ΔCBD  by the SSS congurence theorem

Hence proved


4 0
3 years ago
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