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3 years ago

The diameter of circle A is A. Segment RS B. Segment CA C. Segment CD

Mathematics

2 answers:

stellarik [79]3 years ago

7 0

Answer: RS

Step-by-step explanation:

V125BC [204]3 years ago

4 0

The diameter of circle A is <span>A. Segment RS.
A diameter is a segment, it should include center of a circle (A) and should have 2 points on the circle. </span>

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HELP WITH ONE MANY OR NONE SOLUTION PROBLEM

alexgriva [62]

There would only be on answer as this is not a quadratic equation. Please mark Brainliest!!!

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3 years ago

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Please solve with explanation I’ve been asking all day (this is not a multiple choice question)

Anastaziya [24]

Answer:

a) $SA = 522.9~cm^2$

b) $V_{cone} = 670.2~cm^3$

c) $V_{empty} = 1340.4~cm^3$

Step-by-step explanation:

For a cone,

$SA = \pi r (L + r)$

where L = slant height

$L = \sqrt{r^2 + h^2}$

We have r = 8 cm; h = 10 cm

$L = \sqrt{(8~cm)^2 + (10~cm)^2}$

$L = \sqrt{164~cm^2}$

$SA = (\pi)(8~cm)(\sqrt{164~cm^2} + 8~cm)$

$SA = 522.9~cm^2$

$V_{cone} = \dfrac{1}{3}\pi r^2 h$

$V_{cone} = \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{cone} = 670.2~cm^3$

$V_{cylinder} = \pi r^2 h$

empty space = volume of cylinder - volume of cone

$V_{empty} = V_{cylinder} - V_{cone}$

$V_{empty} = \pi r^2 h - \dfrac{1}{3}\pi r^2 h$

$V_{empty} = (\pi)(8~cm)^2(10~cm) - \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{empty} = 1340.4~cm^3$

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3 years ago

Find the length of rhombus whose diogonals are length 12 cm and 16 cm

grin007 [14]

In ΔAOD (AD) =(AO)

2+(OD)

2 =36+64 AD

2 =100

AD= 100

=10cm

Rhombus has all sides equal

AB = AD = BC = CD = 10 cm

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3 years ago

Lois is riding in the country. after riding her bike at a constant speed for 4 hours, she finds that she traveled a distance of

never [62]

She is going 13 miles an hour or 52 divided by 4

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3 years ago

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Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago