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koban [17]
3 years ago
5

Square roots of 263 by division method ​

Mathematics
1 answer:
tamaranim1 [39]3 years ago
8 0

Answer:

no whole number

Step-by-step explanation:

263 is a prime number, so it can be divided only by 1 and 263. So, no whole number will be square root for 263.

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Step-by-step explanation:

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3 years ago
Between which two integers does square root of 76 lie?
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Answer is B, because 8*8=64 and 9*9=81. 76 lies between 64 and 81.
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the table below shows the function of f determine the value of f(3) that will lead to an average rate of change of 19 over the i
3241004551 [841]

ANSWER

f(3) =  - 25

EXPLANATION

We want to determine the value of f(3) that will lead to an average rate of change of 19 over the interval [3, 5].

The average rate of change of f(x) over the interval [a,b]:

=  \frac{f(b) - f(a)}{b - a}

If the average rate of change over the interval [3, 5] is 19, then;

\frac{f(5) - f(3)}{5 - 3}  = 19

From the to table f(5)=13

\frac{13 - f(3)}{2}  = 19

13 - f(3) = 19 \times 2

13 - f(3) = 38

- f(3) = 38 - 13

- f(3) = 25

f(3) =  - 25

7 0
2 years ago
If 1*2= 5, 3*5= 13, and 2*7= 16, what is 6*8?
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They Multiplied The Second Number By Two Then Added The First Number To It.
5 0
3 years ago
Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \triangle ADC \cong \triangle BDC (LL)

5) \overline{AC} \cong \overline{BC} (CPCTC)

<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

8 0
1 year ago
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