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3 years ago

5

the ratio of dogs to cats at a local animal shelter is 9:3 if there was a total of 33 cats at the shelter how many of them were

dogs

2 answers:

wel3 years ago

8 0

So for every 9 dogs, there are 3 cats. So if I multiply 3 by 11, to get 33. I will also have to multiply 9 by 11.

masya89 [10]3 years ago

6 0

Ee3/5 as a rational number can

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What is the equation, in slope-intercept form, of the line that is perpendicular to the line

NeTakaya

Answer:

y = -x - 4

Step-by-step explanation:

First find the slope

y - 4 = (x - 6)

y - 4 = x - 6

y = x - 6 + 4

y = x - 2 (y = mx + C)

Slope m = 1

It is said that the line is perpendicular to a point ( -2, -2)

If two lines are perpendicular, their slope will be negative reciprocal

The negative reciprocal of slope = 1 is -1

Using a slope intercept form as requested by the question

y = mx + C

Inserting the values given

(-2, -2)

We are using point slope form

y - y_1 = m ( x - x_1)

x_1 = -2

y_1 = -2

m = -1

Insert the values

y - ( -2)) = -1( x - (-2))

y + 2 = -1 ( x + 2)---- point slope form

But we are requested to give the answer in slope intercept form

y = mx + C

We have to open the bracket

y + 2 = -1(x + 2)

y + 2 = -x - 2

y = -x - 2 - 2

y = -x - 4 ( slope - intercept form)

4 0

3 years ago

Solve the system: x + 2 = y ,x2 − 10 = y

kondaur [170]

Answer:

If you wanted the intersecting points there is only one and It's (12, 14)

Hope that could help.

6 0

3 years ago

What is 26.50 rounded to the nearest hundredth

Pavlova-9 [17]

Well,

26.50

In this case, the hundredths place is the 0.

Since there are no more digits after that, 26.50 must already be rounded to the nearest hundredth.

26.50 is the final answer.

6 0

3 years ago

Help!!! Please Answer

adell [148]

Answer:

d sorry if I'm wrong...

6 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago