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3 years ago

Please help meee please

Mathematics

1 answer:

alex41 [277]3 years ago

8 0

Answer:

The answer is Rad 169 or 13 .

Step-by-step explanation:

Hope it helps , good luck .

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What is the area of a square rug that measures 6 feet on a side? 3 square feet 24 square feet 12 square feet 36 square feet

QveST [7]

Well we know the area of a square is l*w, or s^2. So we can use s^2 where s=6 in your case. Now 6^2 is 6*6, which is 36.

So the area of a square rug that measures 6 feet is 36 feet

Answer= 36 feet

4 0

3 years ago

Read 2 more answers

What is the simplified value of the exponential expression 27^1/3?

Kamila [148]

27^1/3= a number you multiply by itself three times to get 27

Cubic root of 27 is 3

= 3

3 0

3 years ago

Janet made 75 more cookies to sell at the

UNO [17]

Answer:

200

Step-by-step explanation:

Janet = x + 75

Kathy = x

2x + 75 = 325

2x = 325 - 75

2x = 250

x/Kathy = 125

x + 75/Janet = 125 + 75

= 200

4 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

Kira ran 8 1/5 miles in 69 7/10 minutes. What is Kira’s average time per mile?

Setler79 [48]

Answer:

1 mile = 8.5 minutes

Step-by-step explanation:

Determining the equation:

Kira ran 8 1/5 miles in 69 7/10 minutes.

⇒ Kira = 8 1/5 miles = 69 7/10 minutes

⇒ 41/5 miles = 697/10 minutes

Simplifying both sides:

⇒ 41/5 miles = 697/10 minutes

⇒ 41 miles = 697/2 minutes

Using cross multiplication:

⇒ 41 × 2 miles = 697 minutes

⇒ 82 miles = 697 minutes

Dividing 82 both sides:

⇒ 82 miles = 697 minutes

⇒ 82/82 miles = 697/82 minutes

⇒ 1 mile = 8.5 minutes

8 0

3 years ago

Please help meee please​

Please help meee please