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DedPeter [7]
3 years ago
5

A segment has endpoints A and C. What are two names for the segment?

Mathematics
2 answers:
Makovka662 [10]3 years ago
8 0

AC and CA

Step-by-step explanation:

line \: which \: has \: endpoint \:  \\  \\ A \:  and  \: C  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\ it \: shows \: that \: line \: starts \: from \: \\  \\  A \:  \: and \: ends \: with \: C \\  \\ so \:  \: its \: name \: is \:  \:  \:  \: AC \\  \\ hope \: it \: is \: helpful \: to \: you....

nexus9112 [7]3 years ago
6 0

Answer:

Ac CA is your answer

Step-by-step explanation:

MARK ME AS BRAINLIEST

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3 years ago
Which functions are vertical stretches of the square root function? Check all that apply.
Amiraneli [1.4K]

Answer:

Step-by-step explanation:

f(x) = √16x

f(x) = 3/8 √9x

f(x) = 6/5 √x

5 0
3 years ago
How do I solve 4/9+9/2 and write my answer as a mixed number?
jenyasd209 [6]

9 has factors of 3*3.

2 is prime, so no factors.

from those two denominators, we can get the LCD of hmmm simply their product, namely 18.

\bf \cfrac{4}{9}+\cfrac{9}{2}\implies \stackrel{\textit{using the LCD of 18}}{\cfrac{(2)4+(9)9}{18}}\implies \cfrac{8+81}{18}\implies \cfrac{89}{18}\implies 4\frac{17}{18}

now, recall that, to get the mixed fraction, we divide 89 ÷ 18, the quotient goes up front, the "4", and the remainder is the one atop, the "17".

8 0
4 years ago
AB=diameter of the circle O. OC=radius. Arc AXB is an arc of the circle with centre C. Prove areas of the shaded region are =
alexandr1967 [171]
1.
Draw a circle with center C And radius CA, as shown in the attached picture.

Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=\sqrt{2}R.

these are all shown in the picture.

2.

Area of triangle ABC is 1/2 * 2R * R= R^2

3.

Let the area between arc BXA and chord AB be Y. (the yellow region).

and let G be the shaded region between arcs AB and AXB.

G=1/2(Area circle with center O)-Y
   =\frac{1}{2} \pi R^{2}-Y

To find Y:

Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.

So Area of sector ACB = \frac{1}{4} \pi (\sqrt{2} R)^{2}=\frac{1}{4} \pi*2 R^{2} =\frac{1}{2} \pi R^{2}

Y =area of sector ABC-Area(triangle ABC)=\frac{1}{2} \pi R^{2}- \frac{1}{2}*2R*R=\frac{1}{2} \pi R^{2}- R^{2}


4. 

Finally,

G=\frac{1}{2} \pi R^{2}-Y=\frac{1}{2} \pi R^{2}-(\frac{1}{2} \pi R^{2}- R^{2})=R^{2}

This proves that the 2 shaded regions have equal area.
 

5 0
3 years ago
Help pleaseeeee<br><br> Thanksss
MissTica

Answer:

-y² + 9y² + 9y +2 =?

Step-by-step explanation:

I put the like terms in parenthesis!

-y² + (6y² + 3y²) + (5y + 4y) +3 - 1 ↓

This is now the simplified and combied expression friend, I hope this helped you!

-y² + 9y² + 9y +2 =?

6 0
2 years ago
Read 2 more answers
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