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3 years ago

. (16/4) + (10-3) Plzz I need an answer quickly. I really would appreciate it

Mathematics

1 answer:

musickatia [10]3 years ago

4 0

Answer:

Explanation

(16/4)+(10-3)

4+(10-3)

4+7

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What is the name of the red dot located on the curve of the parabola below?

katovenus [111]

C. Vertex (x,y). The vertex is right in the middle of the quadratic graph.

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1 year ago

Please help, I’ve been stuck on this question for about a hour now.

Hoochie [10]

6x + 5x + (x + 16) + (3x - 1) = 360
11x + x + 16 + 3x - 1
15x + 15 = 360
15x = 345
x = 23

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3 years ago

Dina has a mass of 50 kilograms and is waiting at the top of a ski slope that’s 5.0 meters high. What is her potential energy

Bess [88]

Answer:

Step-by-step explanation:

Formula for potential energy is PE = m*g*h

where

m is the mass, in kg

g is the acceleration due to gravity, in meters per second squared, and

h is the height, in meters

The problem gives m = 50, g = 9.8, and h = 5. We plug that into the formula and get out answer. So:

$PE=mgh\\PE=(50)(9.8)(5)\\PE=2450$

Correct answer is C.

3 0

3 years ago

Read 2 more answers

What are the value too<br> F(-3)=<br> F(-1)=<br> F(3)=

bagirrra123 [75]

Answer:

f(-3)=-2,5

f(-1)=-1,5

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2 years ago

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$

Hence, $F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1$

Finally; the general solution to the equation is:

$\mathbf{3x^2y^3+2xy^4=C}$

5 0

2 years ago