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mamaluj [8]
3 years ago
14

What is the smallest subset of the square root of 100

Mathematics
1 answer:
aev [14]3 years ago
3 0
=2x5
=10
If you have a certain type of calculator it could answer that question easy

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Am trying to find the value of r<br> 1/r + 2/1-r = 4/r^2
Mashcka [7]
<span>1/r + 2/1-r = 4/r^2

1-r+2r/r(1-r)=4/r^2
(1+r)/r(1-r)=4/r^2   cancle r both side

1+r/1-r=4/r

cross multiply 
r+r^2=4-4r
r^2+4r+r-4=0
r^2+5r-4=0
r^2+4r+r-4=0
solve it for r factor it...



</span>
6 0
3 years ago
How many squares are in this figure in all?
Tanya [424]

Answer:

explanation

Step-by-step explanation:

there

are

22

squares

all

together

6 0
3 years ago
Read 2 more answers
Brainliest for the first solution
mr Goodwill [35]

Q1  Solution:

x = 3 or x = -1

Step-by-step explanation:

x²-2x-3 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -2 and their product -3. By trial and error the two numbers are found to be; -3 and 1. The next step is to split the middle term by substituting it with the above two numbers found;

x²+x-3x-3  =  0

x(x+1)-3(x+1)  = 0

(x-3)(x+1) = 0

Finally we apply the zero Product Property :

If ab = 0 then a  = 0 or b  = 0

This implies;

x-3= 0 or x+1 = 0

x = 3 or x = -1 are the solutions to x²-2x-3 = 0

Q2  Solution:

x  = -1/2 or x  =  3

Step-by-step explanation:

2x²-5x-3  =0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -5 and their product 2(-3)=-6. By trial and error the two numbers are found to be; -6 and 1. The next step is to split the middle term by substituting it with the above two numbers found;

2x²-6x+x-3  = 0

2x(x-3)+1(x-3) = 0

(2x+1)(x-3) = 0

2x+1 = 0 or x-3 =  0

2x = -1 or x =  3

x  = -1/2 or x  =  3 are the solutions of the given quadratic equation.

Q3 Soution:

x = 4 or x = 3

Step-by-step explanation:

x²-7x = -12

x²-7x+12 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -7 and their product 12. By trial and error the two numbers are found to be; -4 and -3. The next step is to split the middle term by substituting it with the above two numbers found;

x²-4x-3x+12 = 0

x(x-4)-3(x-4)  = 0

(x-4)(x-3) = 0

x-4 = 0 or x-3 = 0

x = 4 or x = 3 are the solutions of the given quadratic equation.

Q4:

x = -2/3 or x = 6

Step-by-step explanation:

3x² = 16x+12

3x²-16x-12 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -16 and their product 3(-12)= -36. By trial and error the two numbers are found to be; -18 and 2. The next step is to split the middle term by substituting it with the above two numbers found;

3x²-18x+2x-12 = 0

3x(x-6)+2(x-6) = 0

(3x+2)(x-6) = 0

3x+2 = 0 or x-6 =0

3x = -2 or x = 6

x = -2/3 or x = 6 are the solutions of the given quadratic equation.

Q5:

x = 6 or x = -4

Step-by-step explanation:

x²-2x-24 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -2 and their product -24. By trial and error the two numbers are found to be; -6 and 4. The next step is to split the middle term by substituting it with the above two numbers found;

x²-6x+4x-24 = 0

x(x-6)+4(x-6) = 0

(x-6)(x+4) = 0

x-6 = 0 or x+4 = 0

x = 6 or x = -4 are the solutions to the given quadratic equation.

Q6:

x  = 4/3 or x  = -1

Step-by-step explanation:

3x² = x+4

3x²-x-4 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -1 and their product -12. By trial and error the two numbers are found to be; -4 and 3. The next step is to split the middle term by substituting it with the above two numbers found;

3x²-4x+3x-4 = 0

x(3x-4)+1(3x-4)  =0

(3x-4)(x+1) = 0

3x-4 =0 or x+1 =0

3x  = 4 or x = -1

x  = 4/3 or x  = -1 are the solutions to the given quadratic equation.

5 0
3 years ago
Read 2 more answers
Pls help is due tomorrow and if u help i will give u brainliest
Reil [10]

Answer:

Search it up

Step-by-step equation:

Just search up the website at the bottom of the paper.

6 0
3 years ago
Find the first four terms of this geometric sequence for a1=3 and r=-3
Natali5045456 [20]
a_1=3;\ r=-3\\\\a_2=a_1r\to a_2=3\cdot(-3)=-9\\\\a_3=a_2r\to a_3=-9\cdot(-3)=27\\\\a_4=a_3r\to a_4=27\cdot(-3)=-81\\\\Answer:\{3;-9;\ 27;-81\}
8 0
3 years ago
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