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pishuonlain [190]
3 years ago
15

Solve √(2x-5) - √(x+6) = 0.

Mathematics
1 answer:
nika2105 [10]3 years ago
8 0

Answer:

The solution to this question is 11.

Step-by-step explanation:

Given that

√(2x-5) - √(x+6) = 0

√(2x-5) =√(x+6)  equation (1)

square of the equation (1)

(√(2x-5))^2 =(√(x+6))^2

we know that √a*√a=(√a)^2=a.So

2x-5=x+6          after square.

2x-x=6+5                exchange value

x=11  

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Subtract 8 y^2 − 5 y + 7 from 2 y^2 + 7 y + 1 1 <br><br> The answer is: −6y ^2 +12y+4
lys-0071 [83]

Answer:

\large\boxed{-6y^2+12y+4}

Step-by-step explanation:

(2y^2+7y+11)-(8y^2-5y+7)\\\\=2y^2+7y+11-8y^2-(-5y)-7\\\\=2y^2+7y+11-8y^2+5y-7\qquad\text{combine like terms}\\\\=(2y^2-8y^2)+(7y+5y)+(11-7)\\\\=-6y^2+12y+4

4 0
3 years ago
I need help with this pls
shtirl [24]

Answer:5

Step-by-step explanation:

8 0
2 years ago
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Students are selling pies for a school fundraiser. Customers can buy cherry pies and apple pies. Jennifer sold 12 cherry pies an
Tpy6a [65]

Answer: A cherry pie costs 9 and an apple pie cost 12

Step-by-step explanation:

Let an apple pie be represented by a

Let an cherry pie be represented by c

Jennifer sold 12 cherry pies and 3 apple pies for a total of 144. This can be written as:

12c + 3a= 144

Jacob sold 2 cherry pies and 9 apple pies for a total for 126. This can be written as:

2c + 9a = 126

12c + 3a= 144 ............. i

2c + 9a = 126 ........... ii

Multiply equation i by 2

Multiply equation ii by 12

24c + 6a = 288 ........ iii

24c + 108a = 1512 ...... iv

Subtract iv from iii

-102a = -1224

a = 1224/102

a = 12

An apple pie costs 12

From equation i

12c + 3a= 144

12c + (3×12) = 144

12c + 36 = 144

12c = 144 - 36

12c = 108

c = 108/12

c = 9

A cherry pie costs 9

7 0
3 years ago
This hyperbola is centered at the
Varvara68 [4.7K]

Answer:

The equation is ( x² / 9 ) - ( y² / 7 ) = 1

Step-by-step explanation:

Given the data in question;

hyperbola is centered at the  origin, this means h and k are all equals to 0.

Vertices: (-3,0) and (3,0)

Since y-coordinates are constant, this implies it is a hyperbola with horizontal transverse axis.

h - a = -3

0 - a = -3

a = 3

Foci: (-4,0) and (4,0)

h - c = -4

0 - c = -4

c = 4

we know that, for a hyperbola

c² = a² + b²

so

⇒ ( 4 )² = ( 3 )² + b²

16 = 9 + b²

b² = 16 - 9

b² = 7

So the equation for the hyperbola will be;

⇒ ( (x-h)² / a² ) - ( (y-k)² / b² ) = 1

so we substitute

⇒ ( (x-0)² / 3² ) - ( (y-0)² / 7 ) = 1

⇒ ( x² / 3² ) - ( y² / 7 ) = 1

⇒ ( x² / 9 ) - ( y² / 7 ) = 1

Therefore, The equation is ( x² / 9 ) - ( y² / 7 ) = 1

5 0
2 years ago
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kaheart [24]

Answer:

What help you need from us?????

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