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oee [108]
3 years ago
10

Find the square: (5x - 4y)2 (and please put the process of solving the equation)

Mathematics
1 answer:
satela [25.4K]3 years ago
7 0
(5x-4y)^2
(5x-4y)(5x-4y)

Multiply each term in one bracket in turn by the terms in the other bracket

(5x)(5x) = 25x^2
(5x)(-4y)=-20xy
(-4y)(5x)=-20xy
(-4y)(-4y)=16y^2

Put them together we have
25x^2-20xy-20xy-16y^2
25x^2-40xy-16y^2
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Answer:

Step-by-step explanation:

Let y represent students in the class

55% of y = 78

0.55y= 78

y = 142

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Read 2 more answers
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Gnesinka [82]

Answer:

  • A. 2 < x < 10

Step-by-step explanation:

With two congruent sides the third side is greater against a greater angle.

<u>34 < 36, therefore:</u>

  • 3x - 6 < 24
  • 3x  < 30
  • x < 10

<u>And, the angle should have a positive value:</u>

  • 3x - 6 > 0
  • 3x > 6
  • x > 2

<u>Combined, we get:</u>

  • 2 < x < 10
8 0
3 years ago
10 POINTS! PLEASE HELP!! The illustration shows a compact disc tray that holds five CDs. The radius of each compact disc is 9 cm
MariettaO [177]

We can see that there are 5 CDs, each of radius 9 cm

<u>Area occupied by 1 disc:</u>

Area of a circle = πr²

Area of disc = π(9)²

Area of disc = 3.14 * 81 = 254 cm²

<u>Area occupied by 5 discs:</u>

Area occupied by 5 discs = Area occupied by 1 disc * 5

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Area occupied by 5 discs = 1270 cm²

3 0
3 years ago
Express the quotient of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5C
Lesechka [4]

Answer:

Solution : -\frac{3}{4}-\frac{3}{4}i

Step-by-step explanation:

-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}

=-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right) ÷ 2\sqrt{2}\left(0-1\right)i

= 3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right) ÷ -2\sqrt{2}i

= \frac{3\left(1-i\right)}{\sqrt{2}}÷ 2\sqrt{2}i = -3-3i ÷ 4 = -\frac{3}{4}-\frac{3}{4}i

As you can see your solution is the last option.

3 0
3 years ago
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