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horsena [70]
3 years ago
12

Does anyone have the answer key PDF for Plato (Edmentum) Precalculus A End of Semester Test??

Mathematics
2 answers:
makkiz [27]3 years ago
7 0

Answer:

Step-by-step explanation:

No but if you copy and paste the question to me i can get you the answers :)

Stells [14]3 years ago
5 0

Answer:

I do.

Step-by-step explanation:

Let me know if this works!:)

Download pdf
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What is the best estimated of the total number of students that attended clubs on Friday
s2008m [1.1K]

All you would have to do is round to the nearest 10th number (ex. From 26 to 30) for all the clubs including both boys and girls. Then you add them all together. Rounding makes estimating easier.

8 0
3 years ago
Calculer coefficient de réduction grâce a l'aire
LenKa [72]

Answer:

Comme on a AI = 3 cm et AB = 5 cm, le coefficient de réduction AI / AB est égal à trois cinquièmes, c'est à dire 0,6 . L'aire de AIJ est égale à l'aire de ABC multipliée par k2 .

Step-by-step explanation:

5 0
3 years ago
30 points for anyone that can help please?
statuscvo [17]

Answer:

A) 1.5

B) x^2-23x+42 = (x-2)(x-21)

C) x=2, x=21

Step-by-step explanation:

a) 15 -2x / 3 = 4

15-2x = 12  (Times by three on both sides)

-2x = -3

x = -3/-2 = 3/2 = 1.5

b) x^2-23x+42

x^2-23x+42 = (x-2)(x-21)

c) x^2 -23x +42 = 0

x=2, x=21

(x-2) = 0  ⇒  x = 2

(x-21) = 0  ⇒  x = 21

5 0
3 years ago
Read 2 more answers
Given a term in a geometric sequence and the common ratio, find the explicit formula, a11=5120 ans r=-2
sweet-ann [11.9K]

Answer:

U_n=5(-2)^{n-1}

Step-by-step explanation:

Given a geometric sequence, the nth term of the geometric sequence is obtained by using the formula:

U_n=ar^{n-1}

If the eleventh term, a_{11}=5120

and the common ratio, r=-2.

Then:

5120=a(-2)^{11-1}\\a=\frac{5120}{2^{10}} =5

The explicit formula for the sequence is:

U_n=5(-2)^{n-1}

5 0
3 years ago
Drag each expression to the correct location on the model. Not all will be used
grandymaker [24]

Answer:     \dfrac{x^{2}+2x+1 }{\mathbf {x-1}} \cdot \dfrac{\mathbf {5x^{2} +15x-20} }{7x^{2} +7x}

Step-by-step explanation:

\dfrac{\dfrac{5x^{2} +25x+20}{7x} }{\dfrac{x^{2} +2x+1}{7x^{2} +7x} } =\dfrac{5x^{2} +25x+20}{7x} \times \dfrac{7x^{2} +7x}{x^{2} +2x+1}=\dfrac{5x^{2} +25x+20}{7x} \times \dfrac{7x(x +1)}{(x+1)^{2} }=\\\\=\dfrac{5x^{2} +25x+20}{x+1} =\dfrac{{5(x+1)(x+4)}}{x+1}=5(x+4)

5(x+4)=\dfrac{5(x-1) \cdot (x+4)}{x-1}=\dfrac{5x^{2} +15x-20}{x-1}

<em>Thus, the expressions will be used: (5x² + 15x - 20) and (x + 4).</em>

<em>Let's check:</em>

\dfrac{x^{2}+2x+1 }{\mathbf {x-1}} \cdot \dfrac{\mathbf {5x^{2} +15x-20} }{7x^{2} +7x} =\dfrac{(x+1)^{2} }{x-1}  \cdot \dfrac{5(x-1) \cdot (x+4)}{7x(x+1)} =\\\\=\dfrac{(x+1) \cdot 5 \cdot (x+4)}{7x} =\dfrac{5x^{2} +25x+20}{7x}

5 0
3 years ago
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