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Pie
3 years ago
7

Determine algebraically whether the function is even, odd, or neither. f(x = -5x2 4

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0
What you said doesn't even make sense.

I'll assume you said f(x) = -5x^2 + 4.

This is an even function.

f(-x) = f(x)
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Elis [28]

Answer:

2.35 × 10^(9)

Step-by-step explanation:

2350 million can also be written as;

2350,000,000

Writing that expression in standard form gives;

2.35 × 10^(9)

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How do i solve this problem
Vladimir79 [104]
Solve the following system:
{6 t - 5 s = -4 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{-2 r - 4 s - 4 t = -9 | (equation 3)

Swap equation 1 with equation 3:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Subtract 1/2 × (equation 1) from equation 2:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 1 by -1:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 2 by 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 4 s + 10 t = 1 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Swap equation 2 with equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r - 4 s + 10 t = 1 | (equation 3)
Subtract 4/5 × (equation 2) from equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+(26 t)/5 = 21/5 | (equation 3)

Multiply equation 3 by 5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+26 t = 21 | (equation 3)
Divide equation 3 by 26:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s+0 t = (-115)/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 2 by -5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 2) from equation 1:
{2 r + 0 s+4 t = 25/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 3) from equation 1:
{2 r+0 s+0 t = (-17)/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 1 by 2:
{r+0 s+0 t = (-17)/26 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
v0 r+0 s+t = 21/26 | (equation 3)
Collect results:Answer:  {r = -17/26
               {s = 23/13                        {t = 21/26
7 0
3 years ago
Find the average velocity of the function over the given interval.
snow_lady [41]

Answer:

Average velocity of the function over the given interval

              =  log(\frac{7}{4} ) -2

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given function y = 3/x -2 ...(i)

The average velocity of the function over the given interval

             Average velocity  = \frac{1}{b-a} \int\limits^b_a {(\frac{3}{x} -2)} \, dx

                               =    \frac{1}{7-4} \int\limits^7_4 {(\frac{3}{x} -2)} \, dx

now integrating

                           =   \frac{1}{3}( \int\limits^7_4 {(\frac{3}{x} )} \, dx-2\int\limits^7_4 {1} \, dx )

                           = \frac{1}{3} (3 (log x) - 2 x )_{4} ^{7}

                        =   \frac{1}{3}( (3 (log 7) - 14 )-(3 log 4 -8))

by using formulas

                 log a-log b = log(a/b)

  on simplification , we get                  

                 = \frac{1}{3}( (3 (log 7) -3 log 4 ) - \frac{1}{3} (6)

                = log(\frac{7}{4} ) -2

Average velocity of the function over the given interval

              =  log(\frac{7}{4} ) -2

 

 

6 0
3 years ago
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