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3 years ago

For triangle TRI the following facts are given:

1 answer:

5 0

For the answer to the question above,b.Triangle TAN being an image of triangle TRI is smaller than triangle TRI, and so is a contraction.while,
c.) TA/AN = TR/RI3.3/6 = (3.3 + AR)/86(3.3 + AR) = 8 * 3.3 = 26.43.3 + AR = 26.4/6 = 4.4AR = 4.4 - 3.3 = 1.1AR = 1.1
d.) TA/TN = TR/TI3.3/TN = (3.3 + 1.1)/(TN + 1.6)3.3(TN + 1.6) = 4.4TN3.3TN + 5.28 = 4.4TN4.4TN - 3.3TN = 5.281.1TN = 5.28TN = 5.28/1.1 = 4.8TN = 4.8I hope my answer helped you.

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Referring to the figure, find the missing length.

vivado [14]

Answer:

5 units

Step-by-step explanation:

here;

perpendicular (p)= 3

base (b) = 4

hypotenuse (h) = c= ?

By Pythagorean relationship;

h²=p²+b²

or, h= √(p²+b²)

or, c= √(3²+9²)

or, c= √25

hence, c=5

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2 years ago

A students pay of $21 an hour six dollars more than twice the amount the student as per hour and internship enter in solving equ

Tanya [424]

That makes no sense I think you’re missing some words but the answer might be $7.5 per hour bc 21-6 is 15 and that divided by 2 s 7.5

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3 years ago

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A spa owner gives a survey to all of her customers asking them to rate the quality of the service they received. She then keeps

JulsSmile [24]

I think you forgot to give the options along with the question. I am answering the question based on my research and knowledge. "Some customers who rate their service as high quality return for additional services" is the conclusion that can be drawn from this study. I hope that the answer has actually come to your desired help.

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2 years ago

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(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.

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2 years ago

HELP!!!! I'LL MARK THE BEST ANSWER BRAINLIEST

Aleksandr-060686 [28]

Answer:

x=1

Explain:

y=2x^2−4x+1

dy/dx=4x-4

The line of symmetry will be where the curve turns (due to the nature of the

x^2 graph.

This is also when the gradient of the curve is 0.

Therefore, let

dy/dx=0

This forms an equation such that:

4x−4=0

solve for x,

x=1

and line of symmetry falls on the line

x=1

6 0

2 years ago