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valentina_108 [34]
3 years ago
13

Which statement about the data in the histogram is true?​

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
5 0

Answer: I cant see the picture :')

Step-by-step explanation:

Sorry :'(

gogolik [260]3 years ago
3 0

Answer: C

Step-by-step explanation:

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I’ll mark you brainlist I’ll mark you brainlist show work
Alex787 [66]

Answer:

2=p

Step-by-step explanation:

-5p+7=5p-13

+5p +5p

7=10p-13

+13 +13

20=10p

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2=p

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PilotLPTM [1.2K]
The slope of the line is 1/3
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I am not sure how to slove
notsponge [240]

First .... the limit of the given function is NOT EQUAL to 5 (it diverges so it has no limit).  There is a typo.  The 14 should be positive.

\lim_{x \to\ 7} \bigg(\dfrac{x^2-9x+14}{x-7}\bigg)=5

The precise definition of a limit is:

\text{If for every } \epsilon>0\text{ there exists a }\delta >0\text{ such that}\\|f(x)-L|

Given:

f(x) = \dfrac{x^2-9x+14}{x-7}\\\\L=5\\\\a=7\\\\\\|f(x)-L|

⇒ \epsilon = \delta

When ε = 0.1,     δ = 0.1

When ε = 0.01,   δ = 0.01

5 0
3 years ago
Roberta jogs 3 laps the first day, 5 laps the second day, 7 laps the third day,
iris [78.8K]

Answer:

Sixth day

Step-by-step explanation:

Keep adding 2 until you get to 13

8 0
3 years ago
Please help w this! Its a calculus question! look at the picture for the problem,
neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines x=\pm3\sqrt2 and y=\pm3\sqrt2.

Let R be the region bounded by the line x=3\sqrt2 and the circle x^2+y^2=36 (the rightmost blue region). The right side of the circle can be expressed in terms of x as a function of y:

x^2+y^2=36\implies x=\sqrt{36-y^2}

Then the area of this circular segment is

\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy

=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy

Substitute y=6\sin t, so that \mathrm dy=6\cos t\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18

Then the area of the entire blue region is 4 times this, a total of \boxed{36\pi-72}.

Alternatively, you can compute the area of R in polar coordinates. The line x=3\sqrt2 becomes r=3\sqrt2\sec\theta, while the circle is given by r=6. The two curves intersect at \theta=\pm\dfrac\pi4, so that

\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta

=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18

so again the total area would be 36\pi-72.

Or you can omit using calculus altogether and rely on some basic geometric facts. The region R is a circular segment subtended by a central angle of \dfrac\pi2 radians. Then its area is

\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18

so the total area is, once again, 36\pi-72.

An even simpler way is to subtract the area of the square from the area of the circle.

\pi6^2-(6\sqrt2)^2=36\pi-72

6 0
3 years ago
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