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1 year ago

Please help equation below

Mathematics

1 answer:

qaws [65]1 year ago

7 0

Answer:

3y √21 + 2 y√15

Step-by-step explanation:

To simply, we open up the bracket

We have this as;

√(189y^2) + √60y

= 3y √21 + 2 y√15

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Simplify The following equations

Y_Kistochka [10]

mostly collect like terms
use associative property which is
(a+b)+c=a+(b+c)
also -a+b-c=(-a)+(b)+(-c) so you can move them around
and remember that:
you just use a general rule
x+x=2x
x^2+x^2=2x^2
3xy4xy=7xy
3x+4x^2=3x+4x^2
you can only add like terms( like terms are terms that are same name like x or y are differnt, and like terms have same power exg x^2 and x^3 and x^1/2 and such

I will oly put the naswers because I don't have much time

first one: 2a+3b+2c

second one: remember that -(-6c)=+6c so the answer is c-10a-2b

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2 years ago

Que es 9+10 A.) 45 B.) 19 C.) 21 D.) 24

Vlad [161]

Answer is B. 19! 9+10=19

5 0

1 year ago

Read 2 more answers

Help subtract (xsquared +43x10)-(10xsquared-5x+10 express the answer in standard form

ad-work [718]

Answer:

$\large\boxed{(x^2+43x+10)-(10x^2-5x+10)=-9x^2+48x}$

Step-by-step explanation:

$(x^2+43x+10)-(10x^2-5x+10)\\\\=x^2+43x+10-10x^2-(-5x)-10\\\\=x^2+43x+10-10x^2+5x-10\qquad\text{combine like terms}\\\\=(x^2-10x^2)+(43x+5x)+(10-10)\\\\=-9x^2+48x$

6 0

3 years ago

A rectangular park of length 60 m breadth 50 m encloses w volleyball court of length 18 m and breadth 10 m. Find the area of the

Zanzabum

Given

A rectangular park of length 60 m breadth 50 m encloses with volleyball court of length 18 m and breadth 10 m.

To find:

The area of the park excluding the court at the rate of Rs 110 per square meter.

Solution:

Area of a rectangle is:

$Area=length \times breadth$

Area of whole park is:

$A_1=60 \times 50$

$A_1=3000$

Area of volleyball court is:

$A_2=18 \times 10$

$A_2=180$

Now, the area of the park excluding the court is:

$A=A_1-A_2$

$A=3000-180$

$A=2820$

Therefore, the area of the park excluding the court is 2820 square meter.

4 0

2 years ago

Find the distance from P to l.

ki77a [65]

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$Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}$

4 0

2 years ago