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bearhunter [10]
2 years ago
12

What is the absolute value of 2x + 3 = 11

Mathematics
1 answer:
babunello [35]2 years ago
5 0
The absolute value for 2x+3=11 is x=4
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Ben has square garden that has sides measuring 12 feet. Ben want to increase every side of his garden by x, feet in order to pla
Valentin [98]

Answer:

  • (a) 104 = 4(12+x)
  • (b) x = 14

Step-by-step explanation:

(a) The new side length of Ben's garden is (12+x), since x is the amount of increase above the existing 12 ft side length.

The perimeter is the sum of the lengths of the four equal sides, so is ...

  perimeter = 4 × side length

  104 = 4(12+x)

(b) We can solve this by dividing by 4, then subtracting 12.

  26 = 12 +x . . . . . divide the above equation by 4

  14 = x . . . . . . . . . subtract 12

The solution is x = 14.

4 0
3 years ago
Help me Please I need it !
evablogger [386]

Answer:

38

Step-by-step explanation:

Given that,

x = -4

f(x)=x² - 4x +6

so,

f(-4)= (-4)² - 4 . (-4) +6

       =16 +16 +6

        = 38

4 0
4 years ago
Read 2 more answers
Help please!!!!!!!!!
finlep [7]
Kill me please I'm ugly
4 0
3 years ago
Algebra II Help??? Subtracting Fractions<br><br> (2x)/(y^2-x^2) - (x)/(y-x)
denis23 [38]
Make bottom number same

ok, so remember
(a-b)(a+b)=a²+b²
so

to get from (y-x) to (y²-x²), multiply 2nd fraction by (y+x)

so multiply 2nd fraction by (y+x)/(y+x)
\frac{2x}{y^2-x^2}- \frac{(y+x)(x)}{y^2-x^2}=
\frac{2x-x(x+y)}{y^2-x^2}=
\frac{2x-x^2-xy}{y^2-x^2}
4 0
3 years ago
4^0 + 5^0 + 6^0 = (4+5+6)^0<br> a)True<br> b)False
Yuri [45]

Step-by-step explanation:

false

4^0 + 5^0 + 6^0= 1+1+1=3

(4+5+6)^0= 3⁰=1

1≠3

7 0
3 years ago
Read 2 more answers
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