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Lubov Fominskaja [6]
3 years ago
5

The length of a shadow of the building is 36m. The length of the pole, which is perpendicular to the ground, is 1.9m. The length

of the pole's shadow is 1.52m. What is the height of the building?
Mathematics
1 answer:
attashe74 [19]3 years ago
6 0

Answer:

45

Step-by-step explanation:

We are assuming that the pole's shadow and buildings shadow are proportional so our ratio is

x/36=1.9/1.52

cross multiply

1.52x=68.4

simplify

x=45

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Write a sequence of transformations that maps quadrilateral abcd onto quadrilateral a b c d
kakasveta [241]

It can consist of translations (slides), reflections (flips), and rotations (turns).

I would say a reflection, then slide right, then slide down.

7 0
3 years ago
Hellpp please 20 points for that​
Sergio [31]

Answer:

a) 12cm

b) 360cm²

c) 2160cm cubed

Step-by-step explanation:

a)

you would use Pythagoras theorem  for A

so half of 18 is 9

15² - 9² = 144

the square root of 144 is 12

the height is 12cm

b)

18 x 20 = 360cm²

c)

csa x L

csa = b x h divide by 2

18 x 12 = 216cm²/2 = 108cm²

108xm² x 20cm = 2160cm cubed

8 0
3 years ago
WILL GIVE BRAINLIEST!!!
Charra [1.4K]
I think the answer will be A
5 0
3 years ago
Help!!! I will give brainiest if correct!
maw [93]

Answer:

she would need 7 ounces

Step-by-step explanation:

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
3 years ago
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