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anygoal [31]
3 years ago
7

What is the double of a hundred?

Mathematics
2 answers:
larisa86 [58]3 years ago
8 0

Answer:

200

Step-by-step explanation:

100 * 2 = 200

mote1985 [20]3 years ago
3 0

Answer:

200?

Step-by-step explanation:

Or is it something more complicated

You might be interested in
Solve. (x + 3)^1/2– 1 = x
Natali [406]

Answer:

x=1

Step-by-step explanation:

(x + 3) ^{ \frac{1}{2} }  - 1 = x  \\  \sqrt{x  + 3}   - 1 = x \\  \sqrt{x + 3}  = x + 1 \\ square \: both \: sides \\ x + 3 =  {x}^{2}  + 2x + 1 \\  {x}^{2}  + 2x - x + 1 - 3 = 0  \\  {x}^{2}  + x - 2 = 0  \\ x ^{2}  + 2x - x - 2 = 0  \\ (x^{2}  + 2x) - (x  + 2) = 0 \\ x(x + 2) - 1( x + 2) = 0 \\ (x - 1)(x + 2) = 0 \\ x - 1 = 0 \\ x = 1 \\  \\  \\ x  +  2 = 0 \\ x =  - 2

8 0
3 years ago
Write the polynomial in factored form as a product of linear factors f(r)=r^3-9r^2+17r-9
adelina 88 [10]

Answer:

  f(r) = (x -1)(x -4+√7)(x -4-√7)

Step-by-step explanation:

The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.

The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.

Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:

  r^2 -8r +9 = (r^2 -8r +16) +9 -16 = (r -4)^2 -7

This is zero when ...

  (r -4)^2 = 7

  r -4 = ±√7

  r = 4±√7

Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...

  f(r) = (r -1)(r -4 -√7)(r -4 +√7)

_____

<em>Comment on the graph</em>

I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)

7 0
3 years ago
line m passes through point (-2,-1) and is perpendicular to the graph of y=-2/3x+6. Line n is parallel to line m and passes thro
puteri [66]

Answer:

Hello...... here is a solution :

1 -  

Hello:

the   equation of "m" is : y = ax+b

the slope is a : a×(- 2/3) = -1......( perpendicular to a line : y=-2/3x+6 when the slope is -2/3 )

a = 3/2          

the line " n" that passes through (4, - 3) and  parallel to line "m" :  

when the slope is 3/2 ( same slope )

the equation in slope-intercept form of the line" n" is :

 y – (-3)= (3/2)(x – 4)







7 0
3 years ago
Read 2 more answers
Help please i only got 4 minutes​
Zinaida [17]

Answer:

just set the 2 equations equal to each other. answer is 7, 13/3

Step-by-step explanation:

3 0
2 years ago
Consider the function.
dlinn [17]

Answer:

Statement 3

Step-by-step explanation:

y = -2/3x-24

Make x the subject

y + 24 = -(⅔)x

x = -(3/2)(y + 24)

Intersection variables

f^-1(x) = -(3/2)(x + 24)

Slope: -3/2

No domain restrictions

y-intercept: x = 0

-(3/2)(0+24)

-36

(0,-36)

x-intercepts: y = 0

0 = -(3/2)(x + 24)

x = -24

(-24,0)

Linear function, range is all real values of y

6 0
3 years ago
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