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3 years ago

Please help me!!! some easy algebra

Mathematics

2 answers:

Vadim26 [7]3 years ago

6 0

I think this answer is correct We have read this question like this only

kow [346]3 years ago

4 0

Answer:

5xy-x^2t+2x7+3x^2t= 7xy+2x^2t

Step-by-step explanation:

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What is the result of adding these two equations?<br> 4x – 4y = -2<br> –9x – 4y = -3

White raven [17]

The answer to your question 4x-4y= -2 plus -9x-4y= -3 is -5 because in order to find the result of the two equations would be to add the solutions to both of them together.

7 0

3 years ago

how would you expect the graph of g(x)=f(x-h) to be related to the graph of f(x) when h is a negative number.

Kruka [31]

Answer:

Step-by-step explanation:

The graph of g(x) would be the same as that of f(x) EXCEPT that the graph of g(x) would be translated h units to the left. For example, if you had f(x-[-5]), the graph of f(x) would be translated 5 units to the left.

3 0

3 years ago

Simplify the following rational expression and express in expanded form.

VladimirAG [237]

Answer:

D m = $-\frac{3}{8}$

Step-by-step explanation:

factor the expressions in the denominator and the numerator to simplify the expression:

=> $\frac{2(-2m + 1)(2m - 1)}{2(2m - 1)(8m + 3)}$

=> $-\frac{2m + 1}{8m + 3}$

to make a fraction undefined, the numerator should be 0, thus, we substitute the values of m from the options into the denominator to make the denominator equals to 0:

=> $-\frac{2m + 1}{8(-\frac{3}{8} ) + 3}$ = $-\frac{2m + 1}{-3 + 3}$ = $-\frac{2m + 1}{0}$

in this case, the values of m from option D make the denominator of the fraction equals 0.

6 0

3 years ago

Can you guys tell me which ones are independent varibles or dependent varibles

Mrrafil [7]

the x s are the independent variable so any direct change to them is changing the independent. like the second one has -5 on the x

8 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

3 years ago