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natulia [17]
3 years ago
10

X(x+9)=0 x=? What does x have to be for the answer to be 0?

Mathematics
2 answers:
Nostrana [21]3 years ago
8 0
x(x+9)=0\\
x=0 \vee x=-9
yaroslaw [1]3 years ago
5 0
<em>when we solve the equation it will become
x</em>² + 9x<em>= 0
now here a = 1 , b = 9 and c = 0 
by using quaderatic formula
x = -b +,-</em>√<em>b </em>²<em>- 4ac / 2a     ( 2a is divided by all terms )
by putting values
x = - 9 +,</em><em>- </em>√<em>81 / 2(1)               ( </em>√<em>81 = 9)</em><em>
x = -9 + 9 / 2    and x = -9 -9 /2
x = 0                and x = -18 /2 = -9  
now there are two values for x one is- 9 and other is 0 
so we put -9 as the value of x then answer will be zero..</em>
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Answer:

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Step-by-step explanation:

4 0
3 years ago
Three fourths a number plus 8 is 20 less than the number
VARVARA [1.3K]

Answer:

Equation:  \frac{3}{4}n+8=n-20

Solve for n:  n = 112

Step-by-step explanation:

To first set up the equation, you need to look at the verbal description and translate into numbers and operations:

'three fourths a number' =  \frac{3}{4} n

'plus 8' = + 8

'is' =

'20 less' = - 20

'the number' = n

Put the expressions together:

'three fourths a number plus 8':  \frac{3}{4}n+8

'20 less than the number': n - 20

Set them equal to each other and solve: \frac{3}{4}n+8=n-20

Add 20 to both sides: \frac{3}{4}n+8+20=n-20+20

Subtract \frac{3}{4}n from both sides: \frac{3}{4}n-\frac{3}{4}n+28=n-\frac{3}{4}n

Multiply both sides by \frac{4}{1}: \frac{4}{1}28=\frac{1}{4}n\frac{4}{1}

Solve for n:  n = 112

7 0
3 years ago
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taurus [48]
2. 4/6 , 6/9
3. 2/4 , 3/6
4. 8/10 , 12/15

5. =
6. >
7. >
8. =
9. =
10. <
11. =
12.
6 0
3 years ago
Can anyone please tell me if this is correct? If not help explain me the answer? Thank you
iris [78.8K]
There is on equation to solve for x so I think it is wrong
6 0
3 years ago
which function has real zeros at x = −10 and x = −6? f(x) = x2 16x 60 f(x) = x2 − 16x 60 f(x) = x2 4x 60 f(x) = x2 − 4x 60
LiRa [457]

Answer:

Option A is correct

The function x^2+16x+60 has real zeroes at x =-10 and x =-6

Explanation:

Given: The real zeroes or roots are x = -10, and x = -6

To find the quadratic function of degree 2.

x^2- (\alpha+\beta)x + \alpha\beta =0 where α,β are real roots.   ....[1]

Here, α= -10  and β= -6

Sum of the roots:

α+β =  -10+(-6) = -10-6 = -16

Product of the roots:

αβ = (-10)(-6)= 60

Substitute these value in equation [1] we have;

x^2-(-16)x+60 = x^2+16x+60

Therefore, the quadratic function for the real roots at x =-10 and x =-6 ;

x^2+16x+60

8 0
2 years ago
Read 2 more answers
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