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alex41 [277]
3 years ago
13

Find me midpoint of -9 and 6

Mathematics
2 answers:
Ymorist [56]3 years ago
4 0
The answer I believe is -2. Thank you come again
Alexandra [31]3 years ago
4 0

Answer:

-3/2 or -1.5

Step-by-step explanation:

1) Write out the numbers:

-9, -8, -7, -6, -5, -4, -3, [-2, -1] 0, 1 , 2, 3, 4, 5, 6

2)-2 and -1 are in the middle

3)Add the numbers in the middle and divide by 2

  • Add the 2 numbers: (-2)+(-1)=-3
  • Divide by 2: -3/2
  • Answer: -1.5

                                     

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PLEASE HELP!! look at the image below
Sladkaya [172]

Answer:

A right 4 down 5

Step-by-step explanation:

counting the grid spaces from any point for example R and going to the point R' (R prime) will give you the translation

8 0
3 years ago
FIND THE VALUEOF THE VARIBLE AND FILL THE BLANK.
ludmilkaskok [199]

Answer:

f = 2

g = 8

h = -9

k = 40

m = 1

Step-by-step explanation:

Equation 1:

23f - 17 = 29

Add 17 to both sides. This undoes the -17.

23f = 29 + 17

Add 17 to 29 to get 46.

23f = 46

Divide both sides by 23. This undoes the multiplication by 23.

f = 46/23

Divide 46 by 23 to get 2.

f = 2

Equation 2:

2(3g + 4) = 56

Divide both sides by 2. This undoes the multiplication by 2.

3g + 4 = 56/2

Divide 56 by 2 to get 28.

3g + 4 = 28

Subtract 4 from both sides. This undoes the +4.

3g = 28 - 4

Subtract 4 from 28 to get 24.

3g = 24

Divide both sides by 3. This undoes the multiplication by 3.

g = 24/3

Divide 24 by 3 to get 8.

g = 8

Equation 3:

h + 9 = 0

Subtract 9 from both sides. This undoes the +9.

h = 0 - 9

Any number subtracted from 0 gives its negation.

h = -9

Equation 4

3(k - 8) = 96

Divide both sides by 3. This undoes the multiplication by 3.

k - 8 = 96/3

Divide 96 by 3 to get 32.

k - 8 = 32

Add 8 to both sides. This undoes the -8.

k = 32 + 8

Add 8 to 32 to get 40.

k = 40

Equation 5:

5m - 5 = 0

Add 5 to both sides. This undoes the -5

5m = 0 + 5

Anything plus 0 gives itself.

5m = 5

Divide both sides by 5. This undoes the multiplication by 5

m = 5/5

Anything divided by itself gives you 1.

m = 1

4 0
3 years ago
HELP ASAP!!!! PLSS!!!
Ne4ueva [31]

Answer:

9/32

Step-by-step explanation:

3/4 of 3/8 is 0.28125 or 9/32

3 0
3 years ago
Read 2 more answers
Solve 169x^2-42.25=0 by using square roots?<br><br> NEED HELP!!!
jekas [21]

Answer:

x = 1/2 , x = -1/2

Step-by-step explanation:

Given equation;

169x² - 42.25

Find:

Solution of equation using square root

Computation:

169x² - 42.25

We know that 13² = 169 and 6.5² = 42.25

So,

[13x]² - [6.5]²

a² - b² = (a + b)(a - b)

So,

[13x]² - [6.5]² = [13x + 6.5][13x - 6.5]

So,

13x + 6.5 = 0 , 13x - 6.5 = 0

x = 6.5 / 13 , x = -6.5 / 13

x = 1/2 , x = -1/2

5 0
3 years ago
Someone please help !! I don’t know what I’m doing with this !!
dimulka [17.4K]

Answer:

  a) d(sinh(f(x)))/dx = cosh(f(x))·df(x)/dx

  b) d(cosh(f(x))/dx = sinh(f(x))·df(x)/dx

  c) d(tanh(f(x))/dx = sech(f(x))²·df(x)/dx

  d) d(sech(4x+2))/dx = -4sech(4x+2)tanh(4x+2)

Step-by-step explanation:

To do these, you need to be familiar with the derivatives of hyperbolic functions and with the chain rule.

The chain rule tells you that ...

  (f(g(x)))' = f'(g(x))g'(x) . . . . where the prime indicates the derivative

The attached table tells you the derivatives of the hyperbolic trig functions, so you can answer the first three easily.

__

a) sinh(u)' = sinh'(u)·u' = cosh(u)·u'

For u = f(x), this becomes ...

  sinh(f(x))' = cosh(f(x))·f'(x)

__

b) After the same pattern as in (a), ...

  cosh(f(x))' = sinh(f(x))·f'(x)

__

c) Similarly, ...

  tanh(f(x))' = sech(f(x))²·f'(x)

__

d) For this one, we need the derivative of sech(x) = 1/cosh(x). The power rule applies, so we have ...

  sech(x)' = (cosh(x)^-1)' = -1/cosh(x)²·cosh'(x) = -sinh(x)/cosh(x)²

  sech(x)' = -sech(x)·tanh(x) . . . . . basic formula

Now, we will use this as above.

  sech(4x+2)' = -sech(4x+2)·tanh(4x+2)·(4x+2)'

  sech(4x+2)' = -4·sech(4x+2)·tanh(4x+2)

_____

Here we have used the "prime" notation rather than d( )/dx to indicate the derivative with respect to x. You need to use the notation expected by your grader.

__

<em>Additional comment on notation</em>

Some places we have used fun(x)' and others we have used fun'(x). These are essentially interchangeable when the argument is x. When the argument is some function of x, we mean fun(u)' to be the derivative of the function after it has been evaluated with u as an argument. We mean fun'(u) to be the derivative of the function, which is then evaluated with u as an argument. This distinction makes it possible to write the chain rule as ...

  f(u)' = f'(u)u'

without getting involved in infinite recursion.

7 0
3 years ago
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