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Lemur [1.5K]
2 years ago
14

Jaden made a pot of chili with 64 ounces of ground beef and 2 tablespoons of chili powder. How many pounds of ground beef per ta

blespoon of chili powder did he use?
Mathematics
2 answers:
seropon [69]2 years ago
5 0
32 tablespoons of ground beef
lesya [120]2 years ago
3 0

Answer: 32 ounces

Step-by-step explanation: 64 divide 2 = 32 ounces

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A stadium has a shape of a circle with a diameter 105 meters. Lamp posts will be planted on the circumference of the stadium, wi
grandymaker [24]

Circumference = 105*pi

about 329.8672286269

and that divided by 6 is 54.9778714378, which is about 55

3 0
2 years ago
Complete the equation describing how x and y are related.
shusha [124]

Answer:

y=-5.5x+-5

Step-by-step explanation:

<em>the slope is -5.5</em>

<em>the y-intercept is what the y-axis is when the x-axis is zero</em>

<em>therefore the y-intercept is -5</em>

hope this helps

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7 0
2 years ago
Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 9t + 9 cot(t/2), [π/4, 7π/4]
agasfer [191]

Answer:

the absolute maximum value is 89.96 and

the absolute minimum value is 23.173

Step-by-step explanation:

Here we have cotangent given by the following relation;

cot \theta =\frac{1 }{tan \theta} so that the expression becomes

f(t) = 9t +9/tan(t/2)

Therefore, to look for the point of local extremum, we differentiate, the expression as follows;

f'(t) = \frac{\mathrm{d} \left (9t +9/tan(t/2)  \right )}{\mathrm{d} t} = \frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2}

Equating to 0 and solving gives

\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2} = 0

t=\frac{4\pi n_1 +\pi }{2} ; t = \frac{4\pi n_2 -\pi }{2}

Where n_i is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively

Or we have by chain rule

f'(t) = 9 -(9/2)csc²(t/2)

Equating to zero gives

9 -(9/2)csc²(t/2) = 0

csc²(t/2)  = 2

csc(t/2) = ±√2

The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or

in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2

We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;

f(x) for x = π/4, π/2, 3π/2, 7π/2

f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965

f(π/2) = 9·π/2 +9/tan(π/4) = 23.137

f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412

f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96

Therefore the absolute maximum value = 89.96 and

the absolute minimum value = 23.173.

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3 years ago
Which table is a linear function please help!
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Sorry<br><br> sorryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy
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ITS OKAY dont owrry
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