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3 years ago

Which statements describe a parallelogram that must be a square? Select each correct answer.

Mathematics

2 answers:

Oxana [17]3 years ago

8 0

That would be B, C and E.

A and D could be a rhombus.

DiKsa [7]3 years ago

4 0

Answer:

(D) and (E)

Step-by-step explanation:

From the properties of square, we know that all the four sides of the square are always congruent, the diagonals of the square are always equal and the four angles of the square are equal and are of the measure 90°.

Therefore, from the options, D and E are the correct ones because the sides of square are congruent and the angles are also congruent but the diagonals are not perpendicular.

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Answer:

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Step-by-step explanation:

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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

3 years ago

Use the information given to enter an equation in standard form.

Marysya12 [62]

Answer:

The equation in the standard form of the line through the given points is;

y = 3x - 17

Step-by-step explanation:

We start by getting the slope of the line;

m = (y2-y1)/(x2-x1)

m = (7-10)/(8-9) = -3/-1 = 3

The stands equation form is;

y = mx + b

m is slope, b is the y-intercept

so we have

y = 3x + b

To get b, we substitute the coordinates of any of the points

10 = 3(9) + b

10 = 27 + b

b = 10-27

b = -17

So the equation is;

y = 3x - 17

7 0

2 years ago