All categories

3 years ago

Please help me........

Mathematics

2 answers:

NeX [460]3 years ago

7 0

Answer:

Step-by-step explanation:

j because of the alzemiers algorithim

ikadub [295]3 years ago

4 0

Answer:

Select F

Step-by-step explanation:

G is right triangle, not isoceles

H is also true

J is not true

You might be interested in

You and your best friend go out to dinner. The server was so good that you want to leave a 20% tip. Your food bill was $25.50. W

lys-0071 [83]

Answer:

$5.10

Step-by-step explanation:

We need to find 20% of 25.50

Multiply 20% * 25.50

Change to a decimal

.20 *25.50

5.10

6 0

3 years ago

(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

What is the answer to 2X +1=-15

Over [174]

2X +1=-15
Subtract 1 from both sides
2x=-16
Divide 2 on both sides
Final Answer: x=-8

5 0

3 years ago

Read 2 more answers

I think it’s b but I’m not sure

laila [671]

No, its D

Just look at the rounds mentioned and subtract the scores from higher round with lower round.

Look at A: round 2 score - round 1 score = -2?
-3 -1 = -4 change, not -2 change so it is wrong

Look at B: round 3 score - round 1 score =-1?
-2-1 =-3 change, not -1 change so it is wrong

Look at C: round 3 score - round 2 score =-1?
-2 -(-3) = 1 change, not -1 change so it is wrong

Look at D: round 3 score - round 1 score =-3?
-2-1 = -3 change, matches with -3 so it is correct.

3 0

2 years ago

15. The length of each edge of a regular tetrahedron,

grigory [225]

Slant height of tetrahedron is=6.53cm

Volume of the tetrahedron is=60.35 $\mathrm{cm}^{3}$