Is 128 a square number

Which rule deacribes the

Marysya12 [62]

Answer:

A. ${T_{-6,-2}}\circ {r_{x-axis}}(x,y)$

Step-by-step explanation:

We have that, ΔABC is transformed to get ΔA''B''C''.

We see that the following transformations are applied:

1. Reflection across x-axis i.e. flipped across x-axis.

Now, ΔABC is reflected across x-axis along the line AC to get ΔA'B'C'.

2. Translated 2 units down i.e. shifted 2 units down and and then translated 6 units to the left i.e. shifted 6 units to the left.

So, ΔA'B'C' is translated 2 units downwards and 6 units to the left to get ΔA''B''C''.

Hence, the sequence of transformations is Reflection across x-axis and then Translation of 2 units down and 6 units left.

6 0

3 years ago

If 15 men can do a piece of work in 30 days, in how many days can 10 men complete the same work?

Sav [38]

Answer:

45

Step-by-step explanation:

n/15=30

therefore n=450

hence n/10 = 45

3 0

2 years ago

Read 2 more answers

A square-shaped patio has an area of 256ft squared. What are the dimensions of the patio?

olga nikolaevna [1]

Answer:

all sides are 16 ft

Step-by-step explanation:

If the side length is x, the area is x²=256, hence x = √256 = 16

8 0

2 years ago

A plumber has a 2.5 metre long copper pipe

Anastaziya [24]

We have an object measured in Meters, and we want to cut sections of it off in Centimeters, a different unit of measurement.

Because we're subtracting sections of the pipe, we want to make the units the same, this will make our calculations easier.

1 Meter = 100 Centimeters, so, 2.5 Meters = 250 Centimeters

We're cutting ( 60 Cm + 35 Cm + 90 Cm ) off, which totals 185 Cm.

250 Cm Pipe - 185 Cm Cuts = 65 Cm Pipe Left

8 0

3 years ago

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) Probability of distance at most 100m, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

Probability of distance at most 200:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

Probability of distance between 100 and 200:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

Distance exceeds the mean distance by more than 2σ:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

3 years ago