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Aleksandr [31]
3 years ago
6

The weight of an elephant is 1000 times the weight of a cat. If the elephant weighs 14,000 pounds how many pounds does the cat w

eigh
Mathematics
2 answers:
ohaa [14]3 years ago
5 0
To solve this problem, all you have to do is 14,000/1,000, which is 14, because the elephant weighs 14,000 lbs and the cat is 1/1000 of that weight. The cat weighs 14 pounds.
Ad libitum [116K]3 years ago
4 0
Weight of the elephant= 1,000x
x= weight of the cat
14,000=1,000x
x=14 pounds is what the cat weighs. 
Check: 14*1,000= 14,000
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Which row operation will trangularize this matrix?
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\left[\begin{array}{ccc}1&0&1\\0&0&1\\2&0&1\end{array}\right|\left\begin{array}{ccc}1\\6\\10\end{array}\right] \xrightarrow{-2R_1+R_3}\left[\begin{array}{ccc}1&0&1\\0&0&1\\0&0&-1\end{array}\right|\left\begin{array}{ccc}1\\6\\8\end{array}\right]\\\\Answer:\ C.\ -2R_1+R_3

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2 years ago
Read 2 more answers
As a result of a transfer from one school to another, a boy has to walk one
Softa [21]

Answer:

Let the distance of school and house is x k m

Then time taken at speed 2.5 km\hr=

2.5

x

hr

And time taken at speed 4 km\hr=

4

x

hr

∴

2.5

x

−

4

x

=

60

12

⇒

10

4x−2.5x

=

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1

⇒x=3 km

Then the distance of school and house is 3 k m

may this help you

6 0
2 years ago
Find the length of UT.
otez555 [7]

Answer: A. 32

Concept:

Here, we need to know the idea of the intersecting chord theorem and segment addition postulate.

The<u> intersecting chord theorem </u>states that when two chords intersect at a point, P, the product of their respective partial segments is equal.

The<u> Segment Addition Postulate</u> states that given 2 points A and C, a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.

If you are still confused, please refer to the attachment below for a graphical explanation.

Solve:

<u>Given information</u>

CW = 12

TW = 14

VW = 2x + 5

UW = 2x + 2

<u>Given expression deducted from intersecting chord theorem</u>

CW · VW = TW · UW

<u>Substitute values into the expression</u>

(12) · (2x + 5) = (14) · (2x + 2)

<u>Expand parentheses and apply the distributive property</u>

24x + 60 = 28x + 28

<u>Subtract 14x on both sides</u>

24x + 60 - 24x = 28x + 28 - 24x

60 = 4x + 28

<u>Subtract 28 on both sides</u>

60 - 28 = 4x + 28 - 28

32 = 4x

<u>Divide 4 on both sides</u>

32 / 4 = 4x / 4

x = 8

<u>Given expression deducted from the segment addition postulate</u>

UT = UW + TW

<u>Substitute values into the expression</u>

UT = 2x + 2 + 14

<u>Substitute x value into the expression</u>

UT = 2 (8) + 2 + 14

UT = 16 + 2 + 14

<u>Combine like terms</u>

UT = 32

Hope this helps!! :)

Please let me know if you have any questions

5 0
2 years ago
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