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yarga [219]
2 years ago
15

The formula W=0.86g^2-0.01g+5.9 models the weight, W, in ounces, for a cat who is fed g grams per day of a special food. Accordi

ng to the formula, how much will a cat weigh when it is fed 13 grams of the special food per day? Round to the nearest hundredth.
Mathematics
1 answer:
AfilCa [17]2 years ago
7 0

Weight of cat when 13 grams food is fed is 151.11 ounces.

<h3>Explain equation</h3>

Equations are logical assertions in mathematics that have two algebraic expressions on either side of an equals (=) sign.  LHS = RHS (left hand side = right hand side) appears in all mathematical equations. You can solve equations to determine an unknown variable's value, which corresponds to an unknown quantity.

Given, the formula for weight,

W = 0.86g^{2} - 0.01g + 5.9

    = 0.86(13^{2} ) - 0.01*13 + 5.9

    = 0.86*169 - 0.13 + 5.9

    = 145.34 - 0.13 + 5.9

    = 151.11 ounces

Therefore, the weight of the cat will be 151.11 ounces.

To know more about equations, visit:

brainly.com/question/25307036

#SPJ1

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Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
3 years ago
A croissant shop produces two products: bear claws (B) and almond filled croissants (C). Each bear claw requires 6 oz of flour,
Kruka [31]

Answer:

letter A: B = 400; C = 1000; Max Z = $380

Step-by-step explanation:

bear claws (B) need 6 oz of flour (f), 1 oz of yeast (y), 2 ts of paste (p)

croissants (C) need 3 oz of flour (f), 1 oz of yeast (y), 4 ts of paste (p)

putting that information in equations, we have:

B = 6f + y + 2p

C = 3f + y + 4p

The total number of resources (R) are:

R = 6600f + 1400y + 4800p

Let's call M our total profit, "k1" the number of B produced and "k2" the number of C produced.

So, we can state that:

M = k1*0.2 + k2*0.3

The number of resources R2 will demand is calculated like this:

R2 = k1*B + k2*C = (6k1+3k2)f + (k1+k2)y + (2k1+4k2)p

using R2 and R, we can make some inequations:

6k1+3k2 <= 6600 -> 2k1+k2 <= 2200

k1+k2 <= 1400

2k1+4k2 <= 4800 -> k1+2k2 <= 2400

if we try to maximize k2 (as it worths more), we will have k1 = 0 and k2 = 1200 (limited by p), but looking at the resources R, we will still have resources to use (f and y). Looking at B and C expressions, we see that removing one C gives enough 'p' to make 2 B, which is a good trade (as 2B worths 0.4 cents, and 1C worths 0.3 cents). we have 200 'y' remaining, so doing this 200 times give us k1 = 400 and k2 = 1000, and the only resource remaining will be some of 'f'.

calculating the profit M, we have:

M = 400*0.2 + 1000*0.3 = 380$

the right answer is letter A.

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3 years ago
Find the domain of f(x) = x+5/ 2-x
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I don’t know I am stuck on thisss
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Which statement is true? Please help me
amid [387]
The second option is true: 3/6= 4/8
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3 years ago
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steposvetlana [31]

Answer:

1/4 of the answer

Step-by-step explanation:

6 0
3 years ago
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