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Anni [7]
3 years ago
10

Determine the three-decimal digit approximation of the number v17.

Mathematics
1 answer:
Tanzania [10]3 years ago
4 0

The three decimal fight approximation of the number v17= 4.123

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ASAP Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 24 f
Nitella [24]

Answer:

D. 6 feet.

Step-by-step explanation:

The third longest side of ABCD is 12 feet long and the third longest side of EFGH is 18 feet long.

As they are similar the corresponding sides are in the same ratio so

12/18 =  x/9 where x is the 4th side of ABCD.

2/3 = x/9

3x = 18

x = 6 feet.

3 0
3 years ago
Please Help...........
klasskru [66]
I believe the answer might be:
x = 28/19

Steps:
Expand
3(5 - x)/2 - 4(3 + 2x)/5

-3x + 15 - 8x + 12
------------ -----------
2 5

Find the LCD for the equation above:
(15-3x) * 5 - (12+8x) * 2
------------- --------------
10 10

Now we combine them since the denominators are equal:
5(-3+15) - 2(8x+12)
--------------------------
10

We expand above:
-31x+51
-----------
10
6 0
3 years ago
The first term of a geometric sequence is 300 and the common ratio is 2 What is the 7th term of the sequence
xxMikexx [17]

Step-by-step explanation:

s1 = 300

s2 = s1 × 2 = 300 × 2 = 600

s3 = s2 × 2 = s1 × 2² = 1200

sn = sn-1 × 2 = s1 × 2^(n-1)

s7 = 300 × 2⁶ = 300 × 64 = 19,200

3 0
2 years ago
$8000 at 4% compounded monthly for 9 years
quester [9]
4% of $8000 is $320 x 9 = $2880
The answer is $2880
4 0
3 years ago
Simplify this equation <br> Thanks a lot!!!
den301095 [7]

Answer:

-1

Step-by-step explanation:

We need to simplify the given expression . The given expression is ,

\rm\implies ( 2 \sqrt6 +5)^{2n-1} ( 2\sqrt6 - 5)^{2n-1}

Here we can see that the power of the both exponent is same that is (2n+1) . Recall the property of exponents ,

\bf\implies a^m \times b^m = (ab)^m

Using this property , we have ,

\rm\implies ( 2 \sqrt6 +5)^{2n-1} ( 2\sqrt6 - 5)^{2n-1}

This can be written as ,

\rm\implies\{( 2 \sqrt6 +5) ( 2\sqrt6 - 5)\}^{2n-1}

Simplifying using ( a+b)(a-b) = a² - b² ,

\rm\implies\{( 2 \sqrt6)^2 -5^2 \}^{2n-1}\\\\\rm\implies ( 24 - 25)^{2n-1}

Subtracting the numbers inside the brackets ,

\rm\implies (-1)^{2n - 1 }

Now we know that every odd number is in the form of 2n -1 , where n is any integer. Therefore , the <u>power is odd</u> .

Since the base is (-1) , for even power it is 1 and for odd power it is -1 . Therefore the final answer is ,

\rm\implies\boxed{\quad \red{ -1 }\quad }

<u>Hence </u><u>the</u><u> </u><u>required</u><u> answer</u><u> is</u><u> </u><u>(</u><u>-</u><u>1</u><u>)</u><u> </u><u>.</u>

4 0
3 years ago
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