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Juliette [100K]
3 years ago
10

1. Describe the pattern in each sequence. Then find the next two terms of

Mathematics
1 answer:
dalvyx [7]3 years ago
5 0

Answer:

a b c

Step-by-step explanation:

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I need to know a system of equations to solve this please
Snowcat [4.5K]

3.That BC are congruent and that AB and AC are congruent as well

3 0
3 years ago
A. 3/22 <br> B. 3/7<br> C. 8/15<br> D. 8/11
lorasvet [3.4K]
P(Q|R) = P(Q&R)/P(R)
= (3/37)/(7/37)
= 3/7

The appropriate choice is ...
  B. 3/7
6 0
3 years ago
In the xy-plane, the equations x + 2 = 10 y and 3 +6 = x y c represent the same line for some constant c. What is the value of c
aalyn [17]

Answer:

C=30

Step-by-step explanation:

Supposing that we have the lines

x+2=10y

and

3x+6=Yc.

Note that dividing by 3 the second line can be rewritten as

x + 2= Y c/3.  

Remember that a line is written as ax+b=dY, in our case, both lines have a=1 and b=2. Therefore, in orther that the two lines are equal, we need that 10=c/3, hence c=30.

3 0
3 years ago
These are the first six terms of a sequence with a = 2:
SpyIntel [72]

Answer:

<h2>an+1 = 2×7ⁿ</h2>

Step-by-step explanation:

98÷14

=7

686÷98

=7

4 802÷686

=7

33 614÷4 802

=7

Then the common ratio q for this sequence is 7

recursive formula : an+1 = q×an = ?

an= a1 × qⁿ⁻¹

   =2×7ⁿ⁻¹

 

an+1 = q×an

        = 7×(2×7ⁿ⁻¹)

       = 2×7ⁿ

3 0
3 years ago
Read 2 more answers
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0
3 years ago
Read 2 more answers
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