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vitfil [10]
3 years ago
10

In 193, what digit is in the hundreds place?

Mathematics
1 answer:
Georgia [21]3 years ago
7 0
The digit 1 is in the hundreds place
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scoray [572]

Answer:

SEE THE IMAGE ABOVE FOR ANSWER.

4 0
3 years ago
The data shown has a least-squares regression line of y=3.2x−6.4.
Damm [24]

Answer:

1) B. on, 2) D. 19.2, 3) C. 0

Step-by-step explanation:

1) The value of y according to the regression line is:

y = 3.2\cdot (8) - 6.4

y = 19.2

Hence, the point (8,19.2) is <em>on </em>the least-squares regression line.

2) The value of y according to the regression line is 19.2.

3) The residual is the difference between the value from the regression line and the real value. In this case, the residual value is 0.

5 0
3 years ago
Anyone know the answers or like a website with the answers for these problems they are giving me such a hard time I would GREATL
Digiron [165]

Answer:

just add all sides from the letters it says

Step-by-step explanation:

for an example a has a number b has a number c has a number add all of them together. Hope this helps!

3 0
3 years ago
Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
The angle 118 degrees is also equal to
LekaFEV [45]
I don't understand your question
8 0
4 years ago
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