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3 years ago

Write the following inequality in slope-intercept form. −8x + 4y ≥ −52

2 answers:

6 0

Slope-intercept form:
y=mx+b

-8x+4y≥-52
Add 8x for both side
-8x+8x+4y≥-52+8x
4y≥8x-52
Divided 4 for both side
4y/4≥(8x-52)/4
y≥2x-13. Hope it help!

Shkiper50 [21]3 years ago

4 0

-8x+4y<span>≥-52

Add 8x to both sides:
</span>-8x+8x+4y≥8x-52

Now we have:
4y≥8x-52

Divide everything by 4.

Final answer:
y≥4x-13

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Factor 64 – x15. (4 – x3)(16 + 4x3 + x3) (4 – x3)(16 + 4x3 + x6) (4 – x5)(16 + 4x5 + x5) (4 – x5)(16 + 4x5 + x10)

Dimas [21]

(4-x^5) (16+4x^5+x^10)
I hope this helps!

6 0

3 years ago

Read 2 more answers

(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

Quadrilateral P'Q'R'S' is a dilation of the quadrilateral PQRS, with the origin as the center of dilation.

liraira [26]

Answer:

The rule used to create this dilation is 0.5

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers