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2 years ago

6. Find the mean and the median for these data,

Mathematics

1 answer:

Sholpan [36]2 years ago

7 0

Answer:

73+77+85+84+34+115 and the basket ball team got from the game

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3 years ago

Please help !!! brainliest will be given

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3 years ago

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3 years ago

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find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

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2 years ago

Verify this identity: secx-cosx = 1-cos^2x/cosx

Sauron [17]

Step-by-step explanation:

sec x − cos x

Secant is the inverse of cosine:

(1 / cos x) − cos x

Find the common denominator:

(1 / cos x) − (cos² x / cos x)

Subtract:

(1 − cos² x) / cos x

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3 years ago