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3 years ago

The base of this regular right pyramid is a square. What is its surface area?

Mathematics

1 answer:

zubka84 [21]3 years ago

5 0

Answer:

○ A) 21 ft.²

Step-by-step explanation:

${a}^{2} + 2a \sqrt{ \frac{ {a}^{2}}{4} + {h}^{2} } = S. A. \\ \\ {3}^{2} + 2(3) \sqrt{ \frac{ {3}^{2} }{4} + {2}^{2}} = 24 \\ \\ 24 ≈ 21$

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find the coordinates of the image of point Z (5,-1) when it is translated 1 unit to the left and 2 units down. pls help!!

DaniilM [7]

Answer:

$Z^{'} (5 - 1, -1 - 2) = (4, -3)$

Step-by-step explanation:

The formula of the translation can be written as

$x_2 = x_1 - 1$ , because a shift to the left is along the x - axis and it's negative because it's on the left of the graph.

$y_2 = y_1 - 2$ , because a shift up or down is along the y axis.

To compute the new point Z:

5 0

3 years ago

Read 2 more answers

Number 4 only plzzz help asap its due tomorrow

____ [38]

B:
i=43 degrees
ii= 2 minutes
C:
10 degrees per minute
Sorry that's all I got

7 0

3 years ago

Score Frequency

RUDIKE [14]

Scores on an AP Exam at a local high school are recorded in the table. What was the average score?

Answer: D) 4.266 is correct

Explanation:

We are given the below frequency table:

Score Frequency

1 | 0

2 | 3

3 | 4

4 | 16

5 | 22

Let's denote the score by $x$ and frequency by $f$ .

The average formula is:

$Average=\frac{\sum fx}{\sum f}$

$=\frac{(1\times0) + (2\times3)+(3\times4) +(4\times16) + (5\times22)}{0+3+4+16+22}$

$=\frac{0+6+12+64+110}{45}$

$=\frac{192}{45}=4.267$

Therefore, the average score was 4.267

Hence the option D) 4.266

7 0

3 years ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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6 0

1 year ago

Given P(A)=0.5P(A)=0.5, P(B)=0.71P(B)=0.71 and P(A\cap B)=0.415P(A∩B)=0.415, find the value of P(A|B)P(A∣B), rounding to the nea

Marysya12 [62]

Answer:

0.585

Step-by-step explanation:

P(A|B) = P(A∩B) / P(B)

P(A|B) = 0.415 / 0.71

P(A|B) = 0.585

8 0

3 years ago