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3 years ago

Fractions are equal to 12 /18 is

Mathematics

2 answers:

Amanda [17]3 years ago

8 0

Answer:

2/3

Step-by-step explanation:

FrozenT [24]3 years ago

6 0

Answer:

2/3 = 4/6 = 6/9 = 8/12

= 10/15 = 12/18 = 14/21 = 16/24

= 18/27 = 20/30 = 22/33 = 24/36

= 26/39 = 28/42 = 30/45 = 32/48

= 34/51 = 36/54 = 38/57 = 40/60

= 42/63 = 44/66 = 46/69 = 48/72

= 50/75 = 52/78 = 54/81 = 56/84

= 58/87 = 60/90 = 62/93 = 64/96

= 66/99 = 68/102 = 70/105 = 72/108

= 74/111 = 76/114 = 78/117 = 80/120

= 82/123 = 84/126 = 86/129 = 88/132

= 90/135 = 92/138 = 94/141 = 96/144

= 98/147 = 100/150 = 102/153 = 104/156

= 106/159 = 108/162 = 110/165 = 112/168

= 114/171 = 116/174 = 118/177 = 120/180

= 122/183 = 124/186 = 126/189 = 128/192

= 130/195 = 132/198 = 134/201 = 136/204

= 138/207 = 140/210 = 142/213 = 144/216

= 146/219 = 148/222 = 150/225 = 152/228

= 154/231 = 156/234 = 158/237 = 160/240

= 162/243 = 164/246 = 166/249 = 168/252

= 170/255 = 172/258 = 174/261 = 176/264

= 178/267 = 180/270 = 182/273 = 184/276

= 186/279 = 188/282 = 190/285 = 192/288

= 194/291 = 196/294 = 198/297 = 200/300

You might be interested in

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

3 years ago

Isn’t it just 5? Why would it say check all that apply

LenKa [72]

Answer: $25^\frac{1}{2}$

Step-by-step explanation:

25 = 5*5 or -5 * -5

yes, that is correct. However, you forgot another property of roots. Roots are simply exponential fractions.

$\sqrt[n]{x^m}=x^\frac{m}{n}$

Therefore, $\sqrt{25}$ can also be rewritten as: $25^\frac{1}{2}$ . Notice that it cannot be -25 because that would be a negative square root which will give us an imaginary number.