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tresset_1 [31]
4 years ago
5

Which one of the following examples represents a repeating decimal?

Mathematics
1 answer:
Minchanka [31]4 years ago
3 0
The answer is D. 0.6666666.
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consider the linear function f(x) = 2x-4.if the line was translated 3 units up,what would the new equation be?
dexar [7]

Answer:

f(x)=2x-1

Step-by-step explanation:

If we have a linear function of the form  f(x)=ax-b,

  • Vertical translation of c units up means:

f(x) = ax - b + c

  • Vertical translation of d units down means:

f(x) = ax - b - d

  • Horizontal translation of c units right means:

f(x) = a(x-c) - b

  • Horizontal translation of d units left means:

f(x) = a(x+d) - b

These are the horizontal/vertical translation rules. For the function shown, we apply the first rule and it becomes:

f(x)=2x-4\\f(x)=2x-4+3\\f(x)=2x-1

This would be the new equation of the translated function.

5 0
4 years ago
Joan bought a 3-pound bag of shrimp for 19.55. What is the unit price per ounce of shrimp?
sveta [45]

Answer:

$0.407291667 per ounce

Step-by-step explanation: there are 16 ounces in a pound.   16 times 3 is 48.

in order to find the price per ounce, we need to divide the total price by the number of ounces

19.55/48 is 0.407291667

7 0
3 years ago
Read 2 more answers
Please help me with these questions I’m not very good with This topic
Levart [38]
Use an online converter that you may search for on google. com, and learn the method too
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4 years ago
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Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.
Vesnalui [34]

Answer:

<h3>The value of \frac{\partial w}{\partial s} is e^{3t}-18e^{2s+t}</h3><h3>The value of \frac{\partial w}{\partial t} is 3e^{3t}-9e^{2s+t}</h3><h3>The partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial s} is e^{30}-18</h3><h3>The partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial t}=3(e^{30}-3)</h3>

Step-by-step explanation:

Given that the Function point are w=y^3-9x^2y

x=e^s, y=e^t and s = -5, t = 10

<h3>To find \frac{\partial w}{\partial s} and \frac{\partial w}{\partial t}using the appropriate Chain Rule : </h3>

w=y^3-9x^2y  

Substitute the values of x and y in the above equation we get

w=(e^t)^3-9(e^s)^2(e^t)

w=e^{3t}-9e^{2s}.e^t

<h3>Now  partially differentiating w with respect to s by using chain rule we have </h3>

\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)

=e^{3t}-18e^{2s}.(e^t)

=e^{3t}-18e^{2s+t}

<h3>Therefore the value of \frac{\partial w}{\partial s} is e^{3t}-18e^{2s+t}</h3>

w=e^{3t}-9e^{2s}.e^t

<h3>Now  partially differentiating w with respect to t by using chain rule we have </h3>

\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)

=3e^{3t}-9e^{2s+t}

<h3>Therefore the value of \frac{\partial w}{\partial t} is 3e^{3t}-9e^{2s+t}</h3>

Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :

\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}

=e^{3(10}-18e^{2(-5)+10}

=e^{30}-18e^{-10+10}

=e^{30}-18e^0

=e^{30}-18

<h3>Therefore the partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial s} is e^{30}-18</h3>

\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}

=3e^{3(10)}-9e^{2(-5)+10}

=3e^{30}-9e{-10+10}

=3e^{30}-9e{0}

=3e^{30}-9

\frac{\partial w}{\partial t}=3(e^{30}-3)

<h3>Therefore the partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial t}=3(e^{30}-3)</h3>
6 0
3 years ago
Which sequences of transformations confirm the congruence of shape II and shape I?
defon
The answer is -1 and 12
3 0
3 years ago
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