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iren [92.7K]
2 years ago
15

Football practice starts at 5:45 P.M. and lasts for 80 minutes. What time does football practice end?

Mathematics
2 answers:
vagabundo [1.1K]2 years ago
6 0

Answer:

7:05 pm

Step-by-step explanation:

80 mins =1hr and 20 mins

5.45- 6.45= 1hr

6.45+ 20 mins=7:05pm

bija089 [108]2 years ago
4 0

Answer:

7:05

Step-by-step explanation:

We start with 5:45.

Add 60 minutes / an hour to get 6:45.

We have 20 minutes left to add.

Add 15 minutes to get 7:00.

Add 5 more to get 7:05.

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The two squares x and y are mathematically similar. The areas of x and y are 17cm squared and 272cm squared, respectively. The l
Rasek [7]
I would set this up as a a proportion:
\frac{17cm}{5cm}  · \frac{272cm}{x} 
17x = 1360
x = 20 cm
6 0
3 years ago
Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi
stealth61 [152]
Applying our power rule gets us our first derivative,

\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}

simplifying a little bit,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

looking for critical points,

\rm 0=2x^{-2/3}+4x^{1/3}

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

0=2x^{-2/3}\left(1+2x\right)

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

and take our second derivative.

\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}

Looking for inflection points,

\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}

Again, pulling out the smaller power of x, and fractional part,

\rm 0=-\frac43x^{-5/3}\left(1-x\right)

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
8 0
3 years ago
three fruits are randomly selected from a basket containing 3 apples 2 oranges and 5 kiwis. What is the probability that the thr
stira [4]
3/10 is the probability


8 0
3 years ago
Graph the linear inequality: 2x + 3y = 12, by creating a t-chart to plot the points. Shade by
Crazy boy [7]

Answer:

See explanation

Step-by-step explanation:

Assuming the given inequality is 2x+3y\le12

Then the corresponding linear equation is 2x+3y=12

When x=0, we have  2*0+3y=12

\implies 3y=12\implies y=4

When y=0, we have  2x+3*0=12

\implies 2x=12\implies x=6

The T-table is:

<u>x   |   y</u>

0   |   4

6   |   12

We plot this points and draw a solid straight line as shown in the attachment.

Now let us test the origin: (0,0) by plugging x=0 and y=0 into the inequality.

2*0+3*0\le12

0\le12....This is true so we shade the lower half plane as shown in the attachment.

6 0
3 years ago
HELP I NEED HELP ASAP
Volgvan

the answer is b

X is a bigger number and the difference between x and y is 7, so x-y=7. And it says that the sum of two numbers is 31 so x+y=31

7 0
3 years ago
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