I need help what is this diameter and range

What is 4(1-2b)+7b-10

shutvik [7]

Answer:

-6-b

Step-by-step explanation:

4(1-2b)+7b-10

Multiply what's in parantheses

4(1)=4

4(-2b)=-8b

4-8b+7b-10

Combine like terms

4-10 -8b+7b

-6-b

Hope this helps :)

6 0

3 years ago

Triangle ABC is a right triangle. The sides measure AB = 3 and BC = 4. Use the Pythagorean theorem to find CA. A. CA = 25 B. CA

Vesna [10]

Answer:

CA = 5

Step-by-step explanation:

Since the sides are 3 and 4, we recognize this as a 3:4:5 right triangle, with CA = 5. Using the Pythagorean theorem confirms this:

CA² = AB² +BC²

CA² = 3² +4² = 9 +16 = 25

CA = √25

CA = 5

5 0

3 years ago

Read 2 more answers

PLEASEEEEEEEE HELPPPPPPPPP MEEEEEEEEEE

Archy [21]

Answer:

b

Step-by-step explanation:

In general

Given

y = f(x) then y = f(Cx) is a horizontal stretch/ compression in the x- direction

• If C > 1 then compression

• If 0 < C < 1 then stretch

Consider corresponding points on the 2 graphs

(2, 2 ) → (4, 2 )

(4, - 2 ) → (8, - 2 )

Indicating a stretch in the x- direction.

y = f( $\frac{x}{2}$ ) with C = $\frac{1}{2}$ , that is 0 < C < 1

stretches the graph in the x- direction by a factor of 2

Thus

y = f( $\frac{x}{2}$ ) → b

5 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

Pleassee help me asap

earnstyle [38]

Answer:

2 1/2

Step-by-step explanation:

Hope this helps :P

Have a great day!

6 0

3 years ago