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Kruka [31]
3 years ago
6

I need help please!!!Thank you so much! :)​

Mathematics
2 answers:
slamgirl [31]3 years ago
6 0

Answer:

jen’s speed is 2.4 miles per hour

Kamlees speed is 2 miles per hour

jen is walking faster.

Step-by-step explanation:

Rina8888 [55]3 years ago
3 0

Answer:

Jen is going 135 mph

Kamlee is going 134 mph

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The ratio of 6 hours and 48 minutes in simplest form is (I need this quickly)
julsineya [31]
The gcf of both 6 and 48 is 6. So, in simplest form, the ration would be: 1:6 or 1 hour:6 minutes :)
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3 years ago
In triangle DEF measurement angle D = 45 degrees meausrement angle E = 63 degrees and EF = 24 inches. What is DE to the nearest
alexandr1967 [171]
Since the angles add to 180°, angle F is 72°.

Using the law of sines,

DF/sin72° = 24/sin45°
x = 32.28

So, did you just guess at A?

This is just an answer from another website, but it should still work.
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krok68 [10]
The answer is 300 divided by 20 which you would get 15.
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3 years ago
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Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
Find the points at which the graph of the equation has a vertical or horizontal tangent line. 81x^2 + 49y^2 + 1134x − 882y + 396
Andreyy89

Answer:

882y,com hi

Step-by-step explanation:

hi hi hi hi hi hi hi hi hi

8 0
2 years ago
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