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Hoochie [10]
3 years ago
15

Solve the equation : 7(x + 6) + 7x = 9 options 2 5/14, -1 5/14, -2 5/14, 5/14

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
3 0

Answer:

(i) x + 3 = 0

L.H.S. = x + 3

By putting x = 3,

L.H.S. = 3 + 3 = 6 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ii) x + 3 = 0

L.H.S. = x + 3

By putting x = 0,

L.H.S. = 0 + 3 = 3 ≠ R.H.S.

∴ No, the equation is not satisfied.

(iii) x + 3 = 0

L.H.S. = x + 3

By putting x = −3,

L.H.S. = − 3 + 3 = 0 = R.H.S.

∴ Yes, the equation is satisfied.

(iv) x − 7 = 1

L.H.S. = x − 7

By putting x = 7,

L.H.S. = 7 − 7 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(v) x − 7 = 1

L.H.S. = x − 7

By putting x = 8,

L.H.S. = 8 − 7 = 1 = R.H.S.

∴ Yes, the equation is satisfied.

(vi) 5x = 25

L.H.S. = 5x

By putting x = 0,

L.H.S. = 5 × 0 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(vii) 5x = 25

L.H.S. = 5x

By putting x = 5,

L.H.S. = 5 × 5 = 25 = R.H.S.

∴ Yes, the equation is satisfied.

(viii) 5x = 25

L.H.S. = 5x

By putting x = −5,

L.H.S. = 5 × (−5) = −25 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ix) = 2

L.H.S. =  

By putting m = −6,

L. H. S. =  ≠ R.H.S.

∴No, the equation is not satisfied.

(x) = 2

L.H.S. =  

By putting m = 0,

L.H.S. =  ≠ R.H.S.

∴No, the equation is not satisfied.

(xi) = 2

L.H.S. =  

By putting m = 6,

L.H.S. =  = R.H.S.

Step-by-step explanation:

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6 0
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Calculate the total area of the shaded region.
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\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

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7 0
2 years ago
50 Points!
Advocard [28]
Well, 50 points total given, 25 points per user  and 13 bonus for brainliest

anyway


1.
add them equations, the y's will cancel
x+2y=4
<span>3x-2y=4 +</span>
4x+0y=8

4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution


2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
<span>-4x+14y=-6 +</span>
0x+0y=0
0=0
infinite solutions

that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions



3.
multiply first equation by 2 and add to first
4x+2y=-6
<span>1x-2y=-4 +</span>
5x+0y=-10

5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1

x=-2
y=1
(x,y)
(-2,1)


4.
coincident means they are the same line
so

we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16


5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'


EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a


40 tickets, c+a
40=c+a

the equations are
150=3c+5a and
40=c+a

eliminate
multiply 2nd equaton by -3 and ad to first one

150=3c+5a
<span>-120=-3c-3a +</span>
30=0c+2a

30=2a
divide by 2
15=a

sub back
40=c+a
40=c+15
minus 15
25=c

25 children tickets and 15 adult tickets were sold


ANSWERS:

1.
(2,1) is solution

2.
infinite solutions

3.
(-2,1)

4.
B=10 and C=16

5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
the equations are
150=3c+5a and
40=c+a
25 children tickets and 15 adult tickets were sold
7 0
3 years ago
Read 2 more answers
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