Use the Babylonian method to estimate the square root of 107 to the nearest tenth. 10.2 10.3 10.5 10.4

Mathematics

1 answer:

Delicious77 [7]2 years ago

7 0

I think the answer is B 10.3. Hopefully this helped!! ( and hope I got it correct ) =“)

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Algebraic models grade 12 math college please write the answers without explaining thank you.

Zinaida [17]

From the problem :

$6^4\times6^{-5}$

In multiplying expressions with the same bases, the exponent will be added accordingly.

For example :

$a^m\times a^n=a^{m+n}$

the exponent of a are m and n, and the product will be a raised to the sum of m and n.

Applying this to the problem, we have :

$6^4\times6^{-5}=6^{4-5}=6^{-1}$

The answer is d. 6^-1

6 0

8 months ago

Hey! Can someone check my answers?

Andrew [12]

The greatest common factor is the biggest number taken from the values.

Q1. The answer is <span>A. 5y^6
</span>
$20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}$
Since $x^{a}* x^{b} =x^{a+b}$ , then $x^{9}= x^{3}* x^{6}$

Back to our expression:
$4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+ 5y ^{6}*1=5 y^{6} (4y ^{3} +1)$
The greatest common factor is thus $5 y^{6}$

Q2. The answer is <span>D. 12xy^2
</span>
$12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}$
We will use the rule $x^{a}* x^{b} =x^{a+b}$ to factorise the exponents:
$12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)$
The greatest common factor is thus $12xy^{2}$

3 0

3 years ago

What is wrong with this solution?

Naily [24]

Answer:

They changed 5m into a positive when it should've stayed as a negative until you isolate the variable m at the end, making m actually equal -7.

Step-by-step explanation:

-5m - 16 = 19

-5m = 19 + 16

-5m = 35

m = 35/-5

m = -7