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mina [271]
3 years ago
13

Cantaloupes are on sale at four different grocery stores. Which store offers

Mathematics
1 answer:
dalvyx [7]3 years ago
8 0

Answer:

fruit garden

Step-by-step explanation:Its 60 Cents- answer came from Apex

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3 0
3 years ago
Let h(x)=2x−5 and g(x)=−3x+1.Find h(x)+g(x).? help please
Aleksandr-060686 [28]
This may look a little confusing but all you have to do is plug the equation given for h(x) which is 2x-5 in to the h(x) area in the equation where it says h(x) + g(x).

So far that’s (2x-5) + g(x)

Then,

Do the same with equation given for g(x) which is now,

(2x-5) + (3x+1)

Then solve,

2x - 5 + 3x + 1
2x + 3x- 5 + 1
5x - 4

Therefore the answer is 5x - 4.
8 0
3 years ago
PLEASE HELP 20 POINTS
Murrr4er [49]
Sine = (opposite)/(hypotenuse)

5 is the opposite (opposite of the hypotenuse)
13 is the hypotenuse (longest side)

5/13, or <span>5 ÷ 13 is your answer

hope this helps</span>
6 0
3 years ago
Thomas runs 34 of a mileevery day for 5 days. Howfar has he run total?
STALIN [3.7K]

Guven:

Miles per day = 34 miles

Number of days = 5 days

To find the total distance(miles), we have:

Total distance = distance covered per day x Number per days

= 34 x 5 = 170 miles

Therefore, the total distance he has covered is 170 miles

ANSWER:

170 miles

4 0
1 year ago
Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(
Stells [14]

Answer:

f(x) =  {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x  \rightarrow \: x = sec \: y\\  \frac{dx}{dx}  =  \frac{d(sec \: y)}{dx}  \\ 1 = \frac{d(sec \: y)}{dx} \times  \frac{dy}{dy}  \\ 1 = \frac{d(sec \: y)}{dy} \times  \frac{dy}{dx}  \\1 = tan \: y.sec \: y. \frac{dy}{dx}  \\ \frac{dy}{dx} =  \frac{1}{tan \: y.sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ \sqrt{( {sec}^{2}   \: y - 1}) .sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\   \therefore  \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx}  =  \frac{1}{ |5x |  \sqrt{25 {x }^{2}  - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} }

4 0
2 years ago
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