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eimsori [14]
3 years ago
6

What’s the answer to this question ?????????????

Mathematics
2 answers:
Eduardwww [97]3 years ago
5 0
No, the corresponding angles aren’t congruent. If u add the 135+15+25=175 and the sum of angles must be 180 degrees in all triangles.
ArbitrLikvidat [17]3 years ago
3 0

Answer is is 2 .........

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Does anyone know the answer to this? or how to do it?​
Solnce55 [7]

Answer:

( -2 , 0 ) maximum

Step-by-step explanation:

Watch the graph and point out the coordinates.

if its vertex is less that 0 or negative, then maximum.

With vertex being greater than 0 or positive, its minimum.

7 0
3 years ago
PLZ HLP ASAP WILL MARK BRIANLEIST
alisha [4.7K]
.625 
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5/8 of a can per quart
6 0
3 years ago
Read 2 more answers
How many solutions are there for the system shown below?<br> 2x² + y2 = 33<br> x2 + y + 2y=19
Doss [256]

Answer:2 max because exponet

3 0
3 years ago
Which three lengths could be the lengths of the sides of a triangle?
Montano1993 [528]
C. 7 cm by 7 cm by 43 cm
5 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
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