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2 years ago

What is 12+12 and give good reson why

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

3 0

Answer:

Step-by-step explanation:

you have 12 of something and Are adding 12 to it.

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John & friends brought 4 bags of popcorn and 3 sodas at the movie for a total of 13.50. At the next movie John & friend

scoray [572]

Answer:

The cost of 5 bags of popcorn and 2 sodas is 14.25 units.

Step-by-step explanation:

We are given the following in the question:

Let x units be the cost of a popcorn bag and y units be cost of soda.

The cost of 4 bags of popcorn and 3 sodas is 13.50.

Thus, we can write the equation:

$4x + 3y = 13.50$

The cost of 2 bags of popcorn and 5 sodas is 12.00

Thus, we can write the equation:

$2x+5y=12$

Solving the two equations, we get,

$4x + 3y-(4x+10y) = 13.5 - 24\\\Rightarrow -7y =10.5\\\Rightarrow y = 1.5\\\Rightarrow 2x + 5(1.5) = 12\\\Rightarrow x = 2.25$

Thus, the cost of popcorn bag is 2.25 units and the cost of a soda is 1.5 units.

Cost of 5 bags and 2 sodas

= $=5x+2y\\=5(2.25) + 2(1.5)\\=14.25$

Thus, cost of 5 bags of popcorn and 2 sodas is 14.25 units.

8 0

3 years ago

Which of the following is not a congruence theorem or postulate A. ASA B. SSS C. SAS D. AAA

natka813 [3]

Answer:

D. AAA is not a congruence postulate

Step-by-step explanation:

6 0

2 years ago

What is the value of 3 divided my 64

Naddik [55]

3 divided by 64 is 0.046875

3 0

2 years ago

Help me pls I need it fast

LuckyWell [14K]

Answer:

I don't know does it fold up or anything?

Step-by-step explanation:

3 0

2 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