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Alina [70]
2 years ago
13

Solve for x.

Mathematics
2 answers:
Leto [7]2 years ago
5 0

Answer:

x²+9x=0

x(x+9)=0

either

x=0

or

x=-9

OverLord2011 [107]2 years ago
4 0

Answer:

A. 0, - 9

Step-by-step explanation:

Given

x² + 9x = 0

x (x + 9 ) = 0

Either

x = 0

or,

x + 9 = 0

x = - 9

Hope it will help :)

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What is the effect if two parallel lines are rotated 90° clockwise?
Fed [463]

Answer:

the first one

Step-by-step explanation:

if you have a ladder with no steps and it is leaning on a wall it will be facing up if you carry that ladder sideways it will still be parrallel to the opposite poles

if this confuss you I will answer again

7 0
3 years ago
What is the value of 5^3¡^9
OLEGan [10]

Answer:

125i

Step-by-step explanation:

We want to find

5^3i^9

This can be rewritten as;

5^3i^8\times i

5^3(i^2)^4\times i

Recall that; i^2=-1

5^3(-1)^4\times i

We evaluate now to obtain;

125(1)\times i=125i

6 0
3 years ago
Circumference = ×What is another way to write this?
stepladder [879]

Answer:

2 × pi × radius

..................

8 0
2 years ago
Which expression are equivalent to 1/5 • 1/5 • 1/5 • 1/5
Tresset [83]

Answer:

1/625

Step-by-step explanation:

6 0
3 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
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