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Yuliya22 [10]
3 years ago
7

The quadrilateral ABCD has area of 58 in2 and diagonal AC = 14.5 in. Find the length of diagonal BD if AC ⊥ BD.

Mathematics
2 answers:
MAXImum [283]3 years ago
5 0
We know that

<span>The formula for the area of a quadrilateral with perpendicular diagonals is
</span>A=(0.5)*(D1*D2)
A=58 in²
D1=AC=14.5 in
D2=BD=?

so

D2=2*A/D1-----------> 2*58/14.5-----------> D2=8 in
BD=D2=8 in

the answer is
BD=8 in
mario62 [17]3 years ago
4 0
The <u>correct answer</u> is:

8 in.

Explanation:

The quadrilaterals with perpendicular diagonals are a square and a rhombus.  In both of these, the diagonals are perpendicular bisectors of one another.

Since a square is a rhombus, we will use the formula for the area of a rhombus to solve this problem:

A = (diagonal 1 × diagonal 2)/2

Let diagonal 1 = AC and diagonal 2 = BD.  We know that AC = 14.5 and the area, A, is 58:
58 = (14.5 × BD)/2

Multiply both sides by 2:
58×2 = ((14.5×BD)/2)×2
116 = 14.5×BD

Divide both sides by 14.5:
116/14.5 = 14.5×BD/14.5
8 = BD

The measure of the second diagonal, BD, is 8 inches.
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Answer:

368 cans.

Step-by-step explanation:

Volume of 1 can = π r^2 h.

Here h (height) = 12 and r (radius) = 1/2 * 6 = 3 cm.

So V =  π * 3^2 * 12

= 108π cm^3.

The tank hold 125 liters

=  125,000 cm^3, so:

Number of cans that could be filled = 125000/ 108 π

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2 years ago
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Find the range.<br> 3-97-15-42
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Well to find the range, you need to take the highest number and subtract the lowest from it:

3 - -97

3 + 97

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The perimeter of a rectangular field is 302 yards. If the width of the field is 65 yards, what is its length?​
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Answer:

L= 86 yards

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302= 65(2) + 2(L)

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2 years ago
I need help with #47 to #50 ASAP…. Please help me with it
Ksenya-84 [330]

Step-by-step explanation:

a triangular number n is the sum of all natural numbers <= n.

t1 = 1

t2 = 1+2 = 3

t3 = 1+2+3 = 6

t4 = 1+2+3+4 = 10

...

so,

tn = tn-1 + n

47.

1×8 + 1 = 9 is a square number.

3×8 + 1 = 25 is a square number

6×8 + 1 = 49 is a square number

10×8 + 1 = 81 is a square number

48.

1/3 = 0 remainder 1

3/3 = 1 remainder 0

6/3 = 2 remainder 0

10/3 = 3 remainder 1

15/3 = 5 remainder 0

21/3 = 7 remainder 0

28/3 = 9 remainder 1

so, there seems to be a pattern 1 0 0 1 0 0 1 0 0 1 ...

49.

1/4 = 0 remainder 1

4/4 = 1 remainder 0

9/4 = 2 remainder 1

16/4 = 4 remainder 0

25/4 = 6 remainder 1

36/4 = 9 remainder 0

49/4 = 12 remainder 1

so, there seems to be a pattern 1 0 1 0 1 0 1 0 1 0 1 ...

50.

polygonal numbers is the real name for this.

the formula for dimensions = 5 is

(3n² − n)/2

for dimensions = 6 it is

2n² - n

so, dimensions=5 (and therefore dividing also by 5) we get the remainders

1/5 = 0 remainder 1

5/5 = 1 remainder 0

12/5 = 2 remainder 2

22/5 = 4 remainder 2

35/5 = 7 remainder 0

51/5 = 10 remainder 1

70/5 = 14 remainder 0

92/5 = 18 remainder 2

117/5 = 23 remainder 2

145/5 = 29 remainder 0

here the pattern is 1 0 2 2 0 1 0 2 2 0 1 0 2 2 0 ...

dimensions=6 (and therefore dividing also by 6) we get the remainders

1/6 = 0 remainder 1

6/6 = 1 remainder 0

15/6 = 2 remainder 3

28/6 = 4 remainder 4

45/6 = 7 remainder 3

66/6 = 11 remainder 0

91/6 = 15 remainder 1

120/6 = 20 remainder 0

153/6 = 25 remainder 3

190/6 = 31 remainder 4

231/6 = 38 remainder 3

276/6 = 46 remainder 0

325/6 = 54 remainder 1

here the pattern is 1 0 3 4 3 0 1 0 3 4 3 0 1 0 3 4 3 0 ...

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