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jolli1 [7]
2 years ago
10

R+10<-3(2r-3)+6(r+3)​

Mathematics
1 answer:
MAVERICK [17]2 years ago
8 0

Answer:

{r:r<17}

Step-by-step explanation:

<em>giv</em><em>en</em><em> </em><em>r+10<-3(2r-3)+6(r+3)</em>

change the < to = to solve the equation

thus, r+10=-3( 2r-3 )+6( r+3 )

r+10= -6r +9 +6r +18

r+10= 9+18

r+10=27

r=27-10

r=17

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Ann [662]
No.

4p - 5 = 16

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7 0
3 years ago
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Please help geometry​ questions<br> find X
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Answer:

D

Step-by-step explanation:

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5 0
3 years ago
Determine the number of real solutions for each of the given equations. Equation Number of Solutions y = -3x2 + x + 12 y = 2x2 -
rosijanka [135]

Answer:

Step-by-step explanation:

Our equations are

y = -3x^2 + x + 12\\y = 2x^2 - 6x + 5\\y = x^2 + 7x - 11\\y = -x^2 - 8x - 16\\

Let us understand the term Discriminant of a quadratic equation and its properties

Discriminant is denoted by  D and its formula is

D=b^2-4ac\\

Where

a= the coefficient of the x^{2}

b= the coefficient of x

c = constant term

Properties of D: If D

i) D=0 , One real root

ii) D>0 , Two real roots

iii) D<0 , no real root

Hence in the given quadratic equations , we will find the values of D Discriminant  and evaluate our answer accordingly .

Let us start with

y = -3x^2 + x + 12\\a=-3 , b =1 , c =12\\D=1^2-4*(-3)*(12)\\D=1+144\\D=145\\D>0 \\

Hence we have two real roots for this equation.

y = 2x^2 - 6x + 5\\

y = 2x^2 - 6x + 5\\a=2,b=-6,c=5\\D=(-6)^2-4*2*5\\D=36-40\\D=-4\\D

Hence we do not have any real root for this quadratic

y = x^2 + 7x - 11\\a=1,b=7,-11\\D=7^2-4*1*(-11)\\D=49+44\\D=93\\

Hence D>0 and thus we have two real roots for this equation.

y = -x^2 - 8x - 16\\a=-1,b=-8,c=-16\\D=(-8)^2-4*(-1)*(-16)\\D=64-64\\D=0\\

Hence we have one real root to this quadratic equation.

7 0
2 years ago
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marysya [2.9K]

Answer:

It is four/4 I believe D.4

Step-by-step explanation:

3 0
2 years ago
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mamaluj [8]
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