What is the midpoint between (7 - 3i) and ( 3 + 2i).

HELPPP I WILL BRAINLIST!!

charle [14.2K]

Answer:

Step-by-step explanation:

8 (x -3) +7 = 2x (4 -17)

8(x -3) +7 = 2x (-13) here was the error because the had 13 that is incorrect

8x -24 +7 = -26x

8x -17 = -26x

-17 = -26x -8x

-17 = -34x

-17/-34 = x

1/2 =x

3 0

2 years ago

What number has the same value as 12 tens

torisob [31]

If one ten is ten then just simply multiply 12*10 and its 120 which can be equal to one hundred and two tens

3 0

2 years ago

Read 2 more answers

Please help me answer this for a test im having trouble

Dimas [21]

Answer:

x + 1

y = 9

Step-by-step explanation:

In order to solve this question we need to represent "y "in terms of "x" in the first equation, and the plug in the "y" value in the first equation into the second one. Luckily for us in the first equation it already shows what "y" is equal to in terms of "x" (based on the first equation y = -x + 10). Now we just need to plug in the value that we got instead of "y" in the second equation, and so we get....

y = 7x + 2

(plug in the y value and get the following ….)

-x + 10 = 7x + 2

(now just solve the following equation)

-x + 10 + x = 7x + 2 + x

10 = 8x + 2

10 - 2 = 8x + 2 - 2

8 = 8x

8/8 = 8x/8

1 = x

Now that we know the value of "x", all we need to do now is substitute the value of "x" into any of the equations and we will get the value of "y". So we do the following.....

y = 7x + 2

y = 7(1) + 2

y = 7 + 2

y = 9

3 0

2 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

2 years ago

Is the following relation a function?<br><br> A. Yes<br> B. No

SVETLANKA909090 [29]

Answer:

B. No

Step-by-step explanation:

In order for the relation to be a function, each X value must source exactly one arrow. The value -1 maps to two different values (2, -1), so the relation is NOT A FUNCTION.

3 0

1 year ago