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3 years ago

Tim is a wildlife biologist studying deer parasites

1 answer:

3 0

During a study, he finds that 12 out of 32 deer were affected by parasites. How many of the 1200 deer in the local population would he expect to be affected by parasites?

Answer:

450

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Elisa puts $1,000 into each of 2 bank accounts.

Dafna1 [17]

Answer:

How much money will be in Account A at the end of 3 years? $ ⇒ 1092.73

How much money will be in Account B at the end of 3 years? $112

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Write an equation for the line, given that m= 0 and the y - intercept (0, -2)

stich3 [128]

Remember that the slope intercept formula is:

y = mx + b

m is the slope

b is the y-intercept

so...

m = 0

b = -2

^^^Plug these numbers into formula

y = 0*x - 2

y = 0 - 2

y = -2

Hope this helped!

~Just a girl in love with Shawn Mendes

3 0

3 years ago

Read 2 more answers

The distance dd is 11 mm. What is the perimeter of the figure? Round to the nearest tenth.

Kisachek [45]

Answer:

P = 157.1 mm

Step-by-step explanation:

The perimeter of the figure = Circumference of circle + perimeter of square

We have, d = 11

⇒ Radius = 11 mm

Side of square = 22 cm

So,

$P=2\pi \times 11+4\times 22\\\\P=157.11\ mm\\\\or\\\\P=157.1\ mm$

So, the perimeter of the figure is equal to 157.1 mm.

3 0

3 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

4 years ago

Click on the correct symbol<br> 2,000 lb<br> 1 T

NISA [10]

Answer:

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Step-by-step explanation:

8 0

3 years ago