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2 years ago

5

barnaby has 288 counters in a bag. the counters are red yellow and white. 3/8 of the counters are yellow and white in the ratio

of 1:4. work out how many counters of each colour there are.

1 answer:

postnew [5]2 years ago

8 0

Answer:

idk

Step-by-step explanation:

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What angle does an arc 6.6cm in length subtends at the centre of a circle of radius 14cm. Use π = 22/7)

aniked [119]

Answer:

STEP 1: Find the circumference:

Circumference = 2πr

Circumference = 2π(14) = 28π cm

............................................................................................

STEP 2: Find the length of the arc:

Length of the arc = 36/360 x 28π

Length of the arc = 8.8 cm

.............................................................................................

Answer: The length of the arc is 8.8 cm

............................................................................................

hope it helpssss

Mark it as brilliant answer plzzz

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2 years ago

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Aileen can read 1.5 pages for every page her friend can read. Aileen's mom was very excited and she said to Aileen: "So, if your

dalvyx [7]

No, Aileen's mom is incorrect. Aileen can read 30 pages in the same time period as her friend reads 20 pages.

To get this answer, you multiply 20 by 1.5

20 x 1.5 = 30

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2 years ago

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

7nadin3 [17]

The equation of the cone should be $z=\sqrt{x^2+y^2}$ . Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u\,\vec k$

with $1\le u\le2$ and $0\le v\le2\pi$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k$

Then the integral of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_1^2(-u\cos v\,\vec\imath-u\sin v\,\vec\jmath+u^3\,\vec k)\cdot(u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_1^2(-u^2-u^4)\,\mathrm du\,\mathrm dv$

$=\displaystyle-2\pi\int_1^2(u^2+u^4)\,\mathrm du\,\mathrm dv=\boxed{-\frac{256\pi}{15}}$

3 0

2 years ago

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A person is standing exactly 36 ft from a telephone pole. There is a 30° angle of elevation from the ground to the top of the po

nika2105 [10]

Answer:

20.78feet

Step-by-step explanation:

The question made us to understand that the man is standing and also there is angle of elevation, then we need to draw a right triangle having a base equal to 36 feet with an angle from the base to the top of the pole which is 30 degrees.

tan= opposite side / adjacent side

Let height of the pole =h

Tan(30)= h/36

But tan 30degree= 1/√3

h= 36 × 1/√3

h= 20.78feet

Therefore, the height of the pole= 20.78feet

CHECK THE ATTACHMENT FOR DETAILED FIGURE

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2 years ago

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Prove that (sec 12A-1)/(sec 6A-1)=tan 12A/tan 3A

adelina 88 [10]

Let $x=3A$ . Recall the following identities,

$\cos^2\theta=\dfrac{1+\cos2\theta}2$

$\sin^2\theta=\dfrac{1-\cos2\theta}2$

$\sin2\theta=2\sin\theta\cos\theta$

Now,

$\dfrac{\sec12A-1}{\sec6A-1}=\dfrac{\sec4x-1}{\sec2x-1}$

$=\dfrac{\cos2x(1-\cos4x)}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x(1-\cos^22x)}=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x\sin^22x}=\dfrac{2\cos2x(1+\cos2x)}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x\sin4x}{\cos4x\sin4x}=\dfrac{4\cos2x\cos^2x\tan4x}{\sin4x}$

$=\dfrac{4\cos2x\cos^2x\tan4x}{2\sin2x\cos2x}=\dfrac{2\cos^2x\tan4x}{\sin2x}$

$=\dfrac{2\cos^2x\tan4x}{2\sin x\cos x}=\dfrac{\cos x\tan4x}{\sin x}$

$=\dfrac{\tan4x}{\frac{\sin x}{\cos x}}=\dfrac{\tan4x}{\tan x}=\dfrac{\tan12A}{\tan3A}$

QED

5 0

2 years ago