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andreev551 [17]
3 years ago
9

Help pls i'll give brainliest

Mathematics
2 answers:
romanna [79]3 years ago
6 0

Answer:

They aren't inverse as they intersect with each other more than once.

lilavasa [31]3 years ago
3 0

Answer:

no they aren't inverses bc they arent the same numbers

Step-by-step explanation:

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Five plus three equals
8_murik_8 [283]

Answer: 8

Step-by-step explanation:

5

+

3

8

5 chicken nuggets + 3 more chicken nuggets equals 8

5 0
3 years ago
Read 2 more answers
PLease help me with this!
astraxan [27]

Answer:

1. x2 - 9 > 0

x^2-3^2>0

(x+3)(x-3)>0

(x+3)>0 and (x-3)>0

x>-3       and  x>3

2. x2 - 8x + 12 > 0

   x^2 - 8x  +12>0

   x^2 -2x -6x +12 >0   (-8x is replaced by (-2x) + (-6x) )

   x(x-2) -6(x-2) >0

    (x-6)(x-2)>0

(x-6)>0    and (x-2)>0

    x>6     and     x>2

3. -x2 - 12x - 32 > 0

    -x^2 -12x -32 >0

     x^2 +12x +32 <0

      x^2 +4x +8x +32<0

      x(x+4) +8(x+4)<0

      (x+8)(x+4)<0

(x+8)<0   and  (x+4)<0

x<-8      and    x<-4

4. x2 + 3x - 20 >= 3x + 5

   x^2 +3x -20 >= 3x +5

   x^2 +3x -20 -3x >= 3x +5 -3x

     x^2 -20  >= 5

     x^2 -20 +20  >= 5  +20

     x^2 >=25

     x^2-25 >=0

      (x-5)(x+5)>=0

(x-5)>=0  and (x+5)>=0

  x>=5    and x>=-5

6 0
3 years ago
*i’ll mark brainliest*<br><br> Can someone help me answer #23 A and B. Thanks in advance.
Mila [183]
I would set D = 816000 and then solve
3 0
3 years ago
Find the center and radius for the circles with the following equation x^2+y^2-4x+6y+11=0
Vera_Pavlovna [14]
Rearrange x²+y²-4x+6y=-11 I'll tell you an easy way. take the coefficient of x and divide it by -2 and the coefficient of y and divide it by -2  so -4 ÷-2 = 2           6÷-2= -3 (2,-3) is the center of the circle to know the radius √2² + -3² + -11  =√2  The Radius equals √2
7 0
4 years ago
Find the approximate area between the curve f(x) = -4x² + 32x and on the x-axis on the interval [0,8] using 4 rectangles. Use th
Doss [256]

Split up the interval [0, 8] into 4 equally spaced subintervals:

[0, 2], [2, 4], [4, 6], [6, 8]

Take the right endpoints, which form the arithmetic sequence

r_i=2+\dfrac{8-0}4(i-1)=2i

where 1 ≤ <em>i</em> ≤ 4.

Find the values of the function at these endpoints:

f(r_i)=-4{r_i}^2+32r_i=-16i^2+64i

The area is given approximately by the Riemann sum,

\displaystyle\int_0^8f(x)\,\mathrm dx\approx\sum_{i=1}^4f(r_i)\Delta x_i

where \Delta x_i=\frac{8-0}4=2; so the area is approximately

\displaystyle2\sum_{i=1}^4(-16i^2+64i)=-32\sum_{i=1}^4i^2+128\sum_{i=1}^4i=-32\cdot\frac{4\cdot5\cdot9}6+128\cdot\frac{4\cdot5}2=\boxed{320}

where we use the formulas,

\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2

\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6

6 0
3 years ago
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