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3 years ago

Help !!

Mathematics

2 answers:

KonstantinChe [14]3 years ago

5 0

The bed room would be 2/8

Hope this helps

Nesterboy [21]3 years ago

3 0

You need to add 1/8+2/8+3/8=6/8 so the bedroom is 2/8

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Kenzie has. 4 pages of homework to do. If she can finish 1/3 of a page in one hour how many hours will her homework take?

Maru [420]

Answer:

It will take her 4 1/3 hours

Step-by-step explanation: so you know you needed to add 4 and 1/3 and you turn the 4 into 4/1 and 1/3 and since it adding you have to have the same denominator and you multiply 1x3 and if you do it on the bottom you have to do it on top so you also do 4x3 and 4x3 is 12 and one 1x3 is 3 and so you don't have to have to do it on the other fraction because now they have the same denominator now and now you add 12/3 + 1/3 and you shall get 13/3 and then you divide 13 divided by 3 and you shall get 4 1/3 and that shall be your answer.

hope it helps you!

5 0

3 years ago

For what values of k the relation R:{(k2 , 5), (3k, 6)} is not a function?

Nastasia [14]

Answer:

$k=0$ and $k=3$ .

Step-by-step explanation:

To this relation be a function, the horizontal coordinates can't be equal. So, let's find the value of $k$ that makes those coordinates equal.

$k^{2}=3k\\ k^{2}-3k=0\\k(k-3)=0\\$

Using the null factor property, we have

$k=0\\k-3=0 \implies k=3$

Therefore, the given relation is not a function when $k=0$ and $k=3$ .

7 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

Find the values of the variables and the measures of the angles

Alekssandra [29.7K]

The angles of the triangle are 39°, 51°, and 90°.

Step-by-step explanation:

Step 1:

The sum of all the angles in any triangle is equal to 180°.

We have two of the three angles in terms of x while the third angle is not given directly. Since it is a right-angled triangle, one of the angles equals 90°.

So the angles of the triangle are $3x,$ $4x-1,$ and $90.$