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TEA [102]
2 years ago
9

252.8 x 63.4 please I need help

Mathematics
1 answer:
Lady bird [3.3K]2 years ago
4 0

Answer:

16027.52

Step-by-step explanation:

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Find the first term of the following geometric sequence ----,42,126,378,1,134
sergey [27]
Since it is a geometric sequence each term will be a specific multiple of the term preceding it, called the common ratio...

126/42=378/126=1134/378=3

The sequence is a(n)=a(3)^(n-1) and we know the second term is 42 so:

42=a(3)^(2-1)

42=3a

a=14

So the first term is 14.
8 0
2 years ago
Is the function even, odd, or neither
navik [9.2K]
Which one? There are 2
6 0
2 years ago
Which of the following is the correct factorization of the polynomial below?
serg [7]

Answer:

D but see below.

Step-by-step explanation:

It can be factored if you are allowed to use square roots.

(p + 5√5)(p - 5√5)

Usually, you are not allowed to do that. Usually you must be working with whole numbers or rational well behaved fractions.

I think, given the above comment, the answer is D

4 0
3 years ago
Circle 1: center (8, 5) and radius 6
PolarNik [594]
We have that
<span>Circle 1: center (8, 5) and radius 6
</span><span>Circle 2: center (−2, 1) and radius 2

we know that
the equation of a circle is
(x-h)</span>²+(y-k)²=r²
for the circle 1---------> (x-8)²+(y-5)²=36
for the circle 2---------> (x+2)²+(y-1)²=4

using a graph tool 
see the attached figure

Part A)<span>What transformations can be applied to Circle 1 to prove that the circles are similar?

we know that
r1/r2---------> 6/2------> 3
</span><span>
to prove that the circle 1 and circle 2 are similar, the radius of circle 1 </span>must be divided by 3 and  translate the center of the circle 1  (10) units to the left and (4) units down  
<span>
 the answer part A) is
</span>the radius of circle 1 must be divided by 3 and  translate the center of the circle 1  (10) units to the left and (4) units down  


Part B) <span>What scale factor does the dilation from Circle 1 to Circle 2 have?

the answer Part B) is 
the scale factor is (3/1)</span>

6 0
3 years ago
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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