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3 years ago

Please don't leave half an answer for the points!

Mathematics

1 answer:

natulia [17]3 years ago

8 0

Answer:

T . . . labeling shown below
U
R
S

Step-by-step explanation:

Matrix multiplication of A(a rows, b columns) and B(b rows, c columns) will result in a product matrix of AB(a rows, c columns). This fact can be used to reduce the number of candidates for each product.

For the given factor tiles, the resulting product dimensions will be ...

[2, 3] [3, 2] [2, 2] ⇒ Q, R, S

[2, 3] [3, 3] [3, 2] ⇒ T, U, V

[3, 3] ⇒ W

1. For this 2×3 product, candidates are Q and T. The upper left term for the product in Q is (2)(3) +(1)(7) +(0)(8) = 13. The appropriate choice is T.

2. For this 3×3 product, candidates are U and W. The upper left term for the product in U is 9(2) +1(2) +9(5) +5(1) = 18 +2 +45 +5 = 70. The appropriate choice is U.

3. For this 3×2 product, candidates are R and V. The upper left term for the product in R is 1(4) +2(0) +4(1) = 4 +0 +4 = 8. The appropriate choice is R.

4. For this 2×2 product, there is only one candidate: S.

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Fudgin [204]

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3 years ago

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If an arrow is shot upward on Mars with a speed of 56 m/s, its height in meters t seconds later is given by y = 56t − 1.86t2.

Zina [86]

Answer: (a) (i)50.42 m/s (ii) 51.35 m/s (iii) 52.094 m/s (iv) 52.2614 m/s (v) 52.27814 m/s

(b) 54.14 m/s

Step-by-step explanation:

The average speed over the given time interval [a,b]: = $\frac{y(b)-y(a)}{b-a}$

Given: If an arrow is shot upward on Mars with a speed of 56 m/s, its height in meters t seconds later is given by $y = 56t − 1.86t^2.$

(a)

(i) average speed = $\frac{y(2)-y(1)}{2-1}$

$=\frac{56(2)-1.86(2)^2-(56(1)-1.86(1)^2)}{1}$

$=50.42$ m/s

(ii) average speed = $\frac{y(1.5)-y(1)}{1.5-1}$

$=\frac{56(1.5)-1.86(1.5)^2-(56(1)-1.86(1)^2)}{0.5}$

$=51.35$ m/s

(iii) average speed = $\frac{y(1.1)-y(1)}{1.1-1}$

$=\frac{56(1.1)-1.86(1.1)^2-(56(1)-1.86(1)^2)}{0.1}$

$=52.094$ m/s

(iv) average speed = $\frac{y(1.01)-y(1)}{1.01-01}$

$=\frac{56(1.01)-1.86(1.01)^2-(56(1)-1.86(1)^2)}{0.01}$

$=52.2614
$ m/s

(v) average speed = $\frac{y(1.001)-y(1)}{1.001-01}$

$=\frac{56(1.001)-1.86(1.001)^2-(56(1)-1.86(1)^2)}{0.001}$

$=52.27814
$ m/s

(b) $y(1)=56(1)-1.86(1)^2$

$=56-1.86$

$= 54.14$ m/s

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3 years ago

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ValentinkaMS [17]

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ludmilkaskok [199]

I hope this helps you

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3 years ago

A fire station is to be located along a road of length A, A < q. If fires occur at points uniformly chosen on (0, A), where s

Shtirlitz [24]

Answer:

Step-by-step explanation:

Given that X is uniform in the interval (0,A)

X is continuous since it represents the distance

A fire station is to be located along a road of length A, A < q. If fires occur at points uniformly chosen on (0, A), we should find where the station be located so as to minimize the expected distance from the fire

Distance can be taken as absolute values here as either side is the same.

E(|x-a|] is to be minimum

E(|x-a|)=E(x-a) for 0<x<a

=E(a-x), for a<x<A

Using integral we find this value

$E(|x-a|)=\int\limits^a_0 {x-a} \, dx +=\int\limits^A_a -{x-a} \, dx\\f(a)=\frac{a^2}{2} -\frac{(A-a)^2}{2}$

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$a=\frac{A}{2}$

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4 years ago