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3 years ago

Does anyone know the answer to this?

Mathematics

2 answers:

pishuonlain [190]3 years ago

8 0

Answer:

Rectangle

Step-by-step explanation:

(hope this helps!!)

Dvinal [7]3 years ago

3 0

Your answer would be rectangle:)

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Fine the probability. NEED THE STEPS IF POSSIBLE.

Mekhanik [1.2K]

Answer:

The probability of getting an even number on a number cube is 1/2

The probability of getting the heads on a dime is 1/2

Step-by-step explanation:

A number cube has six numbers

And three of those numbers are even numbers.

Therefore;

Probability = 3/6

= 1/2

A dime has two faces, heads and tails

One of them is heads

Therefore;

Probability = 1/2

5 0

3 years ago

Helppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

dezoksy [38]

answer:

A is your answer...

explanation:

I just divided each number by 4 to get the results..

hope it helps and plz mark brainliest

8 0

2 years ago

Read 2 more answers

(X^2+y^2+x)dx+xydy=0 Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

is exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of x, and dividing this by $N(x,y)=xy$ gives an expression independent of y. If we assume $\mu=\mu(x)$ is a function of x alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by x on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

$\dfrac{\partial(x^2y)}{\partial x}=2xy$

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

The solution F satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

$\dfrac{\partial F}{\partial y}=x^2y$

Integrating both sides of the first equation with respect to x gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to y gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

5 0

3 years ago

HELP!!! ASAP

andrey2020 [161]

It can be read as "six cubed."

It can be read as "six to the power of three."

It has a base of 6.

It is the same as multiplying three factors of 6.

It has an exponent of 3.

Have a Nice Day

4 0

3 years ago

On Monday a Farmer sold 25196 pounds of potatoes on tuesday he sold 18023pounds on Wednesday he sold some more potatoes in all h

arlik [135]

Answer:

estimated sales on Wednesday is 19000 pounds.

Step-by-step explanation:

On Monday, he sold 25196 pounds. Estimated to the nearest thousand that is 25000 pounds.

On Tuesday, he sold 18023 pounds. Estimated to the nearest thousand, that is 18000 pounds

Wednesday's sales is unknown. We designate as x

All in all he sold 62409. Estimated to the nearest thousand, that is 62000

The sales on Monday, plus sales on Tuesday, plus sales on Wednesday, must all sum up to the total sales.

25000 + 18000 + x = 62000

43000 + x = 62000

x = 62000 - 43000 = 19000

therefore estimated sales on Wednesday is 19000 pounds.

3 0

3 years ago