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1 year ago

My grandson thinks he can cook better than any other

1 answer:

4 0

The answer is (B). Employing a comma to separate two independent phrases connected by the conjunction "and," avoids the mistake of the original.

<h3>What are jumbled words?</h3>

The reason was given for Wrong Answer A: Incorrect coordination is present in Option (A). Where a comma is required, a semicolon is used instead.

The reason was given for the Wrong Answer C: The sentence produced by option (C) is illogical. It illogically implies that the grandson's success in winning blue ribbons is due to his belief that he is the fair's top chef.

Justification for Wrong Answer D: In the case of option (D), the sentence is irrational. The initial portion of the line, "To think... fair," and the latter section of the sentence has no connection to one another ("my grandson . . . to prove it.").

Justification for Wrong Answer E: The sentence produced by option (E) is irrational. It is illogical to claim that the grandson's blue ribbons demonstrate that he was "thinking he can cook better... fair."

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Zanzabum

Check the picture below.

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2 years ago

Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

sashaice [31]

Let $R$ be the solid. Then the volume is

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
(by computing the Jacobian)

and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$

7 0

3 years ago

Find the area of sector AOB. Express your answer in terms of pi and rounded to the nearest tenth.

STatiana [176]

63 $\pi$

The sector is $\frac{70}{360}$ ; there are 360 degrees in a circle.
You use the formula $\pi r^{2}$ .
18cm is the radius, and then you square it to get 324.
Multiply 324 by $\frac{70}{360}$
You get 63 $\pi$

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3 years ago

The quotient of a number and (-3) is (-80) Find the number.

USPshnik [31]

Answer:

240

Step-by-step explanation:

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3 years ago

1. The prism-shaped roof has equilateral triangular bases. Create an equation that models the height of one of the roof's triang

Mandarinka [93]

1. Check picture 1. Let the one side of the triangle be a, drop one perpendicular, CD. Then triangle ADB is a right triangle, with hypothenuse a and one side equal to 1/2a. By the Pythegorean theorem, as shown in the picture, the height is $\frac{ \sqrt{3} }{2} a$

2. if a a=25 ft, then the height is $\frac{ \sqrt{3} }{2} a=\frac{ \sqrt{3} }{2} *25= \frac{1.732}{2}*25= 21. 7$ (ft)

3. consider picture 2. Let the length of the roof be l feet.

one side of the prism (the roof) is a rectangle with dimensions a and l, so the area of one side is a*l
the lateral Area of the roof is 3a*l

the area of the equilateral surfaces is $2*( \frac{1}{2} *a* \frac{ \sqrt{3} }{2}a )=\frac{ \sqrt{3} }{2} a^{2}$

so the total area of the roof is $\frac{ \sqrt{3} }{2} a^{2} +3al$

4. The total area was the 2 triangular surfaces + the 3 equal lateral rectangular surfaces. Now instead of 3 lateral triangular surfaces, we have 2.
So the total area found previously will be decreased by al

5. so the area now is $\frac{ \sqrt{3} }{2} a^{2} +2al$

6. now a=25 and l=2a=50

Area= $\frac{ \sqrt{3} }{2} a^{2} +2al=\frac{ \sqrt{3} }{2} * 25^{2}+2*25*50=25 ^{2} (\frac{ \sqrt{3} }{2} +4)=625*4.866$

=3041.3 (ft squared)

7 0

3 years ago

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