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DedPeter [7]
4 years ago
9

(image) 20 point Mathematics​

Mathematics
1 answer:
Sauron [17]4 years ago
4 0

Answer:

A.

Step-by-step explanation:

r=diameter/2

radius=52/2 (Q)

Q =26

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Question 7 of 33
astraxan [27]

Answer:

90°.

Step-by-step explanation:

When two straight lines intersects obliquely then there will be two angle of intersection, one angle is acute and the other one is obtuse.

But, here the two straight lines AB and xy intersects perpendicularly and AB bisects xy.

Therefore, the two angle of intersection are 90° each.

Hence, the angle of intersection in our case will be equal to 90°. (Answer)

7 0
4 years ago
I love you NO H O M O ;)
Pachacha [2.7K]

Answer:

thx for the pts

Step-by-step explanation:

have a great day :)

8 0
3 years ago
Nt Addition and Cogs tru<br> D<br> 13<br> F.
Serga [27]

Answer:

666665555

Step-by-step explanation:

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7 0
3 years ago
The first difference of a sequence is the arithmetic sequence
ad-work [718]

Answer:

a). 1, 3, 5, 7, 9, 11

b). 4, 6, 8, 10, 12, 14

c). 28, 30, 32, 34, 36, 38

Step-by-step explanation:

<u>Solution for question (a):</u>

<u />

The first difference of a sequence is the arithmetic sequence;

2, 4, 6, 8, 10

As clearly shown, the common difference (d) is 2.

This sequence, above, is an arithmetic sequence.

If the first term of an arithmetic sequence (AP) is 1, then the first six terms would be:

1, 3, 5, 7, 9, 11

<u>Solution for question (b):</u>

<u />

If the sum of the first two terms of an AP is ten,

Lets say the first term is x and the second term is x + 2 then;

x + (x + 2) = 10

2x = 10 - 2

x = 4

So the first term is 4 and the six terms of the AP are:

4, 6, 8, 10, 12, 14

<u>Solution for question (c):</u>

<u />

If the fifth term of an AP is 36 then,

To find the nth term in an AP we use the formula;

T_{n}  = a + (n - 1)d  (where T_{n} is the nth term, a is the first term and d is the common difference).

36 = a + (5 - 1)2

36 = a + 8

a = 36 - 8 = 28

So the first term is 28 and the first terms are;

28, 30, 32, 34, 36, 38

3 0
3 years ago
How do you solve this? (#11/#12)
zepelin [54]

\text{If}\ x_1,\ \text{and}\ x_2\ \text{are zeros of a polynomial function, then the function}\\\text{has equation}:\\\\f(x)=a(x-x_1)(x-x_2).\\----------------------------\\\\11.\ \text{We have the zeros}\ 3\ \text{and}\ i:\\\\f(x)=(x-3)(x-i)=(x)(x)+(x)(-i)+(-3)(x)+(-3)(-i)\\\\=x^2-xi-3x+3i=\boxed{x^2-(3+i)x+3i}\\\\12.\ \text{We have the zeros}\ -2\ \text{and}\ 2i:\\\\f(x)=(x-(-2))(x-2i)=(x+2)(x-2i)\\\\=x^2-(2i)^2=x^2-2^2i^2=x^2-4(-1)=\boxed{x^2+4}\\\\\text{used}\\\\(a-b)(a+b)=a^2-b^2\\\\i=\sqrt{-1}\to i^2=-1

3 0
3 years ago
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