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3 years ago

Find the volume of the following square pyramid. geometry

Mathematics

2 answers:

Mkey [24]3 years ago

8 0

Answer:

volume = l×b×h

= 2×2×3

=12

Step-by-step explanation:

Hope it will help

<h2>please mark as a brainlists</h2>

TiliK225 [7]3 years ago

7 0

Answer:

Volume of a pyramid = lwh /3 ==12/3==4

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What is 3%+5%+4565+76+90x6(7+5)?

lawyer [7]

I believe it’s 5,193.0315 I’m not to sure

6 0

3 years ago

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If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that

Ivan

Answer:

$\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

Given :-

• $\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}$

To Prove :-

• $\sf\implies cos2A =\dfrac{\sqrt3}{2}$

Proof :-

We know that ,

$\sf\implies cos2A = 1 - 2sin^2A$

Therefore , here substituting the value of sinA , we have ,

$\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2$

Simplify the whole square ,

$\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}$

Add the numbers in numerator ,

$\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8}$

Multiply it by 2 ,

$\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}$

Take out 2 common from the numerator ,

$\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}$

Simplify ,

$\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}$

Subtract the numbers ,

$\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}$

Simplify,

$\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }$

Hence Proved !

8 0

3 years ago

What is an equation for the linear function whose graph contains the points (−1, −2) and (3, 10)

sladkih [1.3K]

The equation of a line starting from two points is:
$y-y_1=\frac{y_2-y_1}{x_2-x_1} \cdot (x-x_1)$

From the first point you get: x1 = -1, y1 = -2
From the second point you get: x2 = 3, y2 = 10

Replace x1, y1, x2, y2 in the equation of the line and you get:
$y+2=\frac{10+2}{3+1} \cdot (x+1)$
$y+2=\frac{12}{4} \cdot (x+1)$
$y+2=3 \cdot (x+1)$

From this you get the equation of your line:
$y(x)=3x+1$

4 0

3 years ago

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The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

lys-0071 [83]

Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

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4 years ago

Prove that the curve a(t) = (cost, sin 2t, cos 2t) is regular on R and that it self-intersects at (1,0,1). Check the self-inters

alexandr402 [8]

Answer:

The function a (t) is a vector function composed of the component functions $a_ {1} (t) = cost, a_ {2} (t) = sin2t$ and $a_ {3} (t) = cos2t$ . How $a_ {1} (t), a_ {2} (t), a_ {3} (t)$ are infinitely derivable functions in R, so they are regular functions in R.

Now, for $t = 0$ , you have to $a (0) = (cos (0), sin2 (0), cos2 (0)) = (1, 0, 1)$ . How the functions $a_ {1} (t), a_ {2} (t), a_ {3} (t)$ are periodic functions with period $2 \pi,$ the vector function $a (t)$ will take the same point $(1, 0 , 1)$ at $t = 2n\pi, n = 0, 1, 2, 3, ...$ then the vector function is auto-intercepted

Step-by-step explanation:

6 0

3 years ago