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s344n2d4d5 [400]
3 years ago
8

In the figure two semicircles are drawn at the centre of the large semicircle

Mathematics
2 answers:
MakcuM [25]3 years ago
8 0

Answer:

picture is missing.

Step-by-step explanation:

picture is missing

Nitella [24]3 years ago
4 0
Answer
Were is the picture
Explanation
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Find the range of y = -3x + 7 for the domain {-8, -6, -3, -2, 2}
svetlana [45]
The domain is the set of x-values of a function. The range is the set of y-values of a function.

You are told that the domain, or x-values, are -8, -6, -3, -2, and 2. To find the range, you just need to plug in each of the x-values into the function <span>y = -3x + 7 and find the value of y.
1) When x = -8:
</span><span>y = -3x + 7
y = -3(-8) + 7
y = 24 + 7
y = 31

2) When x = -6
</span>y = -3x + 7
y = -3(-6) + 7
y = 18 + 7
y = 25

3) When x = -3
y = -3x + 7
y = -3(-3) + 7
y = 9 + 7
y = 16

4) When x = -2
y = -3x + 7
y = -3(-2) + 7
y = 6 + 7
y = 13

5) When y = 2
y = -3x + 7
y = -3(2) + 7
y = -6 + 7
y = 1

The range is {31, 25, 16, 13, 1}.

-------

Answer: 
{31, 25, 16, 13, 1}
5 0
3 years ago
From the side view, a gymnastics mat forms a right
Natali [406]

Answer:

its b

Step-by-step explanation:

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3 years ago
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Jack has 3/4 cup of Tide left in the bottle. Each load of laundry requires 1/6 cup of Tide. How many loads of laundry can Jack w
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Answer:

4 1/2 loads

Step-by-step explanation:

3/4 / 1/6 = 3/4 * 6 =  18/4 = 4 1/2


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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.

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3 years ago
Line cd has endpoints c(1,4) and d(5,0) line ef has endpoints (4,5) and (2,-1).​
amm1812

Answer:

yo what up

Step-by-step explanation:

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3 years ago
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