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scZoUnD [109]
3 years ago
15

Jonathan took a 35 questions test and

Mathematics
2 answers:
Musya8 [376]3 years ago
6 0

Answer:

22.86%

Step-by-step explanation:

(wrong questions)/total questions * 100(%)= (35-27)/35 * 100(%)=22.857...%

Round to the nearest hundredth: 22.86%

Goshia [24]3 years ago
3 0

Answer:

%38

Step-by-step explanatio35 - 27 =8

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Step-by-step explanation:

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25a-20

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What piece of information is needed to prove<br> the triangles are congruent through ASA?
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Plsss help !!!!!!!!!!!!!!!​
seropon [69]

Problem 37

<h3>Answer: B. 4 & 1/4 gallons</h3>

-------------------

Work Shown:

The 3/4 and 1/2 add up to (3/4)+(1/2) = (3/4)+(2/4) = 5/4 = 1&1/4

The whole part 1 is then added to the other whole parts 1 and 2 to get 1+2+1 = 4

Overall, the grand total is 4 & 1/4

An alternative way is to convert each mixed number to a decimal to get 1&1/2 = 1+1/2 = 1+0.5 = 1.5 and 2&3/4 = 2+3/4 = 2+0.75 = 2.75

Add up those decimals 1.5+2.75 = 4.25

then convert that result to a mixed number: 4.25 = 4+0.25 = 4+1/4 = 4&1/4

================================================

Problem 38

<h3>Answer: C. Cubic feet</h3>

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The units for volume follow the template "Cubic _____" where you'll write "feet", "inches", "centimeters" or whatever unit you are using in the blank.

Some other examples:

* cubic miles

* cubic meters

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3 0
3 years ago
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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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