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e-lub [12.9K]
3 years ago
11

Write the integers that are closest to the the number in the middle to complete the statement.

Mathematics
2 answers:
zaharov [31]3 years ago
8 0

Answer:

8<square root of 73<1

i think this are the integers that are closest to the number in the middle.

ArbitrLikvidat [17]3 years ago
6 0

Answer:

see explanation

Step-by-step explanation:

Consider perfect squares on either side of 73, that is

64 < 73 < 81 , then

\sqrt{64} < \sqrt{73} < \sqrt{81} , that is

8 < \sqrt{73} < 9

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Factorise the following equation: <img src="https://tex.z-dn.net/?f=6x%5E2%2B17x%2B5" id="TexFormula1" title="6x^2+17x+5" alt="6
RSB [31]

Answer:

(3x+1)(2x+5) is the answer

8 0
3 years ago
Find the range of allowable values based on an expected mass of 250 grams for which values are allowed to vary by 5%
PIT_PIT [208]

Answer:

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4 0
2 years ago
What kind of function is 3y-2=x+7
Phoenix [80]
I agree with the answer above
4 0
3 years ago
Given A is between Y and Z YA = 5.5, AZ = 2x, and YZ = 41.5 =, find AZ
morpeh [17]
YZ = YA + AZ
41.5 = 5.5 + AZ
AZ = 41.5 - 5.5
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So, your final answer is AZ = 36

Hope this helps!
4 0
3 years ago
Read 2 more answers
Geometric sequences HELP ASAP!
Pani-rosa [81]

Given:

The table for a geometric sequence.

To find:

The formula for the given sequence and the 10th term of the sequence.

Solution:

In the given geometric sequence, the first term is 1120 and the common ratio is:

r=\dfrac{a_2}{a_1}

r=\dfrac{560}{1120}

r=0.5

The nth term of a geometric sequence is:

a_n=ar^{n-1}

Where a is the first term and r is the common ratio.

Putting a=1120, r=0.5, we get

a_n=1120(0.5)^{n-1}

Therefore, the required formula for the given sequence is a_n=1120(0.5)^{n-1}.

We need to find the 10th term of the given sequence. So, substituting n=10 in the above formula.

a_{10}=1120(0.5)^{10-1}

a_{10}=1120(0.5)^{9}

a_{10}=1120(0.001953125)

a_{10}=2.1875

Therefore, the 10th term of the given sequence is 2.1875.

6 0
2 years ago
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