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laila [671]
2 years ago
12

1 point

Mathematics
1 answer:
Slav-nsk [51]2 years ago
7 0
38 ma)a mamma mamahbahwnnwhshsjhsjsjwjsjsjshsuhs
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Cari owns a horse farm and a horse trailer that can transport up to 8,000 pounds of livestock and tack. She travels with 5 horse
Serggg [28]
5 horses with a combined weight of 6,240 pounds
horse trailer has a limit of 8,000 pounds of horse and tacks.

Find the weight of tack, t, for each horse that Cari can allow.

5t + 6,240 <  8,000

5t + 6,240 < 8,000
    - 6,240    -6,240
5t              < 1,760
<span>÷5                 ÷5    </span>
 t               <  352

Each horse can carry a tack that weighs no more than 352 pounds. 

Any tack that weighs beyond <span>352 </span>pounds is will exceed the maximum carrying capacity of the horse trailer.


4 0
3 years ago
Read 2 more answers
PLEASE HELP ME ASAP?!?(2questions) 60POINTS+BRAINLIEST
Elodia [21]

It says that there was an error with my answer, but I don't know what, so I screenshotted what I typed and attached them as 4 images below:

4 0
2 years ago
Let p = log10 x, q = log10 y, and r = log10 z. (The 10s are base numbers.)
KatRina [158]

Answer:

(p / q^2 √r)

Just make x, y, z into p, q, r.

Very simple!

4 0
3 years ago
What is the quotient of 2/5 divided by 3/7<br><br>A. 15/14<br>B. 14/15<br>C. 35/6<br>D. 6/36​
STALIN [3.7K]

Answer:

= { \tt{ \frac{2}{5}  \div  \frac{3}{7} }} \\  \\  = { \tt{ \frac{2}{5} \times  \frac{7}{3}  }} \\  \\  = { \underline{ \tt{ \:  \:  \frac{14}{15}  \:  \: }}}

<u>Answer</u><u>:</u><u> </u><u>B</u>

8 0
2 years ago
Read 2 more answers
Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = x − ln 8x, [1/2, 2]
insens350 [35]
The given function is 
f(x) =  x - ln(8x), on the interval [1/2, 2].

The derivative of f is
f'(x) = 1 - 1/x
The second derivative is 
f''(x) = 1/x²

A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0  => 1/x = 1  => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079

The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2)  = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.

Answer: 
Minimum value  = 1 - ln(8)
Maximum value = 2 - ln(16)


4 0
3 years ago
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