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arsen [322]
3 years ago
15

1. Find the values of x.

Mathematics
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

x = 9√3

Step-by-step explanation:

x^2 = 18^2 - 9^2

x^2 = 324 - 81

x^2 = 243

x = 9√3

You might be interested in
Solve 3x square - 2 x + 7 x - 5
ale4655 [162]

Answer:

The correct ans is...

3 x square + 5 x - 5

Step-by-step explanation:

U can only subtract or add if u hv the same variable..

here -2x and 7x have the same variable...

so -2x + 7x = 5x

Therefore the ans is....

3 x square + 5 x - 5

Hope this helps....

Have a GOOD DAY !!!!

Pls mark my ans as brainliest

If u mark my ans as brainliest u will get 3 extra points

Sry to say....

U really need to read ur Math TBK

7 0
3 years ago
Read 2 more answers
3a <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3a%20%2B%20a%20%7B%7D%5E%7B2%7D%20%7D%7Ba%7D%20" id="TexFormula1" title="
egoroff_w [7]

Answer:

(3+a)

Step-by-step explanation:

3a + a^2

-------------

a

Factor out an a in the numerator

a(3+a)

-------------

a

Cancel like terms

(3+a)

6 0
3 years ago
Read 2 more answers
TRUE OR FALSE: Two chords in the same circle are congruent if and only if the associated central angles are congruent.
Lostsunrise [7]

Answer:

Step-by-step explanation:

True

7 0
3 years ago
Read 2 more answers
(HELP ME 15 PTS)
kap26 [50]

Answer:   \bold{\dfrac{27x^2y^3}{16}}

<u>Step-by-step explanation:</u>

\dfrac{(3x^2y^5)^3}{(2xy^3)^4}\\\\\\=\dfrac{3^3\cdot x^{2\cdot 3}\cdot y^{5\cdot 3}}{2^4\cdot x^4\cdot y^{3\cdot 4}}\\\\\\=\dfrac{27\cdot x^6\cdot y^{15}}{16\cdot x^4\cdot y^{12}}\\\\\\=\dfrac{27\cdot x^{6-4}\cdot y^{15-12}}{16}\\\\\\=\dfrac{27\cdot x^2\cdot y^3}{16}

8 0
3 years ago
Simplify x^2+xy-yx+y^2 + y^2+yz-zy+z^2 + z^2+zx-xz+x^2
Artemon [7]

Answer:

2x^2 + 2y^2 + 2z^2

Step-by-step explanation:

x^2+xy-yx+y^2 + y^2+yz-zy+z^2 + z^2+zx-xz+x^2

First step is to put all the same ones together

( FYI, xy and yx is the same thing, just like how

2 x 3 and 3 x 2 is the same thing) this time I'll bracket the groups to make it easier on the eyes

(x^2 +x^2) + (xy - xy) +( y^2 + y^2) + (yz - yz) +

(z^2 +z^2) + (zx - zx)

2x^2 + 2y^2 + 2z^2

All the others just cancel themselves out, for example, xy-xy

When anything minus themselves it will become 0

7 0
3 years ago
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