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Illusion [34]
3 years ago
6

Find the value of x. (Note: the line passing through the center of the circle is a diameter)

Mathematics
1 answer:
ikadub [295]3 years ago
8 0

Where's the picture?  

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A sequence is defined by the following: an = 4n – 1 What is a1?
Alexandra [31]

an = 4n – 1

a1  means n=1

a1 = 4(1) -1

a1 = 4-1

a1 =3

Answer: 3

6 0
3 years ago
Read 2 more answers
URGENT:
scoray [572]
I would say the answer would be B

8 0
3 years ago
Read 2 more answers
4x^2-2x<br> ————<br> X^2+2x+1
Len [333]

Answer:

x = 0 or x = 1/2 and x = -1 for the second equation

Step-by-step explanation:

Solve for x:

4 x^2 - 2 x = 0

Factor x and constant terms from the left hand side:

2 x (2 x - 1) = 0

Divide both sides by 2:

x (2 x - 1) = 0

Split into two equations:

x = 0 or 2 x - 1 = 0

Add 1 to both sides:

x = 0 or 2 x = 1

Divide both sides by 2:

Answer: x = 0 or x = 1/2

_______________________________

Solve for x:

x^2 + 2 x + 1 = 0

Write the left hand side as a square:

(x + 1)^2 = 0

Take the square root of both sides:

x + 1 = 0

Subtract 1 from both sides:

Answer: x = -1

4 0
3 years ago
3. The shaded part of the model shows what part of a
Tcecarenko [31]

Answer:

\dfrac{1}{6}

Step-by-step explanation:

From the diagram, the model is divided into 6 equal parts.

The shaded part of the model shows what part of the garden is planted with peas.

Since, 1 part out of 6 is shaded.

The fraction which names the shaded part is therefore: \dfrac{1}{6}

One-Sixth of the garden is planted with peas.

7 0
2 years ago
Describe the relationship between the segments made when secant lines intersect outside a circle. Use the following image as ref
Natali [406]
Let the extensions of secants AD and BC meet at point P.

Let the measure of ar DC be 2a, then m(DAC)=m(DBC)=a, since angles DAC and DBC are both inscribed angles intercepting arc DC.

m(P)=b.

Then triangles APC and BPD are similar, because they have 2 pairs of common angles.

from the similarity of APC and BPD, we can write the ratios:

\frac{AP}{BP} = \frac{PC}{PD} = \frac{AC}{BD}

5 0
3 years ago
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