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nordsb [41]
3 years ago
5

The total amount of fiber (in grams) in a package containing x apples and y oranges is given by the equation 5x + 10y = 110. Is

it possible for the package to contain 15 apples?
Mathematics
2 answers:
garri49 [273]3 years ago
5 0
It is, if you are allowed to have half an orange. If not, then you cannot have 15 apples.

15= x, so you’d plug that into the equation and solve like this:
5(15) + 10y = 110
75 + 10y = 110
Subtract 75 from both sides
10y = 35
Divide ten by both sides
y= 3.5
Flauer [41]3 years ago
3 0
The answer is NO
Because if x=5 the equation is 125+10y=110
In this case y will be a negative
You can’t have a negative oranges
So the answer is NO
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Approximate f by a Taylor polynomial with degree n at the number a. Step 1 The Taylor polynomial with degree n = 3 is T3(x) = f(
MAVERICK [17]

Answer:

If you center the series at x=1

T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)

Where R(x) is the error.

Step-by-step explanation:

From the information given we know that

f(x) = e^{2x^2}

f'(x) = 4x e^{2x^2}   (This comes from the chain rule )

f^{(2)}(x) = 4e^{2x^2} (4x^2+1)   (This comes from the chain rule and the product rule)

f^{(3)}(x) = 16xe^{2x^2}(4x^2 + 3)  (This comes from the chain rule and the product rule)

If you center the series at x=1  then

T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)

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3 0
3 years ago
PLZ HELP !! Which of the points (s, t) is not one of the vertices of the shaded region of the
stira [4]

Answer:

A(0,10)

Step-by-step explanation:

Given the 4 inequalities:

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s ≥ 10 -t (2)

s ≤ 20-t (3)

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Let analyse all 4 possible answer:

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Let substitute this point into (1) we have: 10 ≥ 12 -0.5*0 Wrong

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Let substitute this point into (1) we have: 20 ≥ 12 -0.5*0  True

Let substitute this point into (2) we have: 20 ≥ 10 - 0 True

Let substitute this point into (3) we have: 20 ≤ 20 - 0 True

We choose B as the vertex

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Let substitute this point into (1) we have: 16 ≥ 12 -0.5*4  True

Let substitute this point into (2) we have: 16 ≥ 10 - 4 True

Let substitute this point into (3) we have: 16 ≤ 20 - 4 True

We choose C as the vertex

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Let substitute this point into (1) we have: 4 ≥ 12 -0.5*16 True

Let substitute this point into (2) we have: 4 ≥ 10 - 16 True

Let substitute this point into (3) we have: 4 ≤ 20 - 16 True

We choose B as the vertex

Hence, the point A(0,10)  is not one of the vertices of the shaded region of the  set of inequalities.

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