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3 years ago

7

Find the slope and y-intercept of y=10x

2 answers:

masha68 [24]3 years ago

5 0

Answer:

the slope would be 10

and the y-intercept would be 0

saveliy_v [14]3 years ago

5 0

Slope: 10

y-intercept: (0, 0)

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‼️‼️help would be appreciated‼️

quester [9]

24.

This is because the equation you have presented is that of a circle, formulated like so:

x^2 + y^2 = r^2 or radius squared

With that information, you know the find the square of the presented radius which is 12 and the actual radius.

The radius is half of the diameter, which makes the diameter 12 x 2 or 24.

Hope this helps!

4 0

3 years ago

What is the multiplicative inverse of 16?

fomenos

Answer:1/16

Step-by-step explanation:

4 0

3 years ago

Simplify: (6x-4) (9x-2)

Vinil7 [7]

Step-by-step explanation:

Hey, there!!

While simplifying the algebraic expression, you must be very careful about the signs.

Here,

(6x-4) (9x-2)

Multiply 2nd expression by 1st expression,

= 6x ( 9x - 2) -4 ( 9x - 2)

$= (54 {x}^{2} - 12x) - (36x - 8)$

Opening brackets,

$= 54 {x}^{2} - 12x - 36x + 8$

Simplifying them we get,

$= 54 {x}^{2} - 48x + 8$

Therefore, the answer is 54x^2 - 48x +8.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

5 0

3 years ago

In a previous exercise we formulated a model for learning in the form of the differential equation dP dt = k(M − P) where P(t) m

GalinKa [24]

Answer:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$

Step-by-step explanation:

For this case we aasume the following differential equation:

$\frac{dP}{dt}= k(M-P)$

Is a separable differential equation so we can do the following procedure:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$

8 0

3 years ago

What is the solution of this system of linear equations? Y= 2x-3; x= y-3

Ivanshal [37]

X=y-3,yer that’s the answer

4 0

4 years ago

Read 2 more answers