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kobusy [5.1K]
3 years ago
6

Which fraction is closest to 1/2? A:1/6 B:3/8 C:3/4 D:-1/2

Mathematics
1 answer:
Mandarinka [93]3 years ago
5 0
B. 3/8 is the answer




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Which expression is equivalent to 3 7/8-6 1/4
Serhud [2]

get a common denominator which is 24

3 7/8 = 3 21/24

6 /4 = 6 6 /24

3 21/24 - 6 6 /24

big number minus small number take sign of larger

6 6/24

- 3 21/24

-------------


5  30/24

3  21/24

--------------

2  9/24


2 3/8

the sign of the larger is negative

-2 3/8


7 0
3 years ago
Someone please help me out i don´t understand this question <br> ill give you 21 points
suter [353]

Answer:

0.67

Step-by-step explanation:

Formula: k = y/x

k = 2/3

k = 0.66666666666

k = 4/6

k= 0.66666666666

k = 6/9

k = 0.66666666666

k = 8/12

Round 0.67

Hence, answer = 0.67

[RevyBreeze]

8 0
3 years ago
Read 2 more answers
Evaluate x(-y + z) for x = 5, y = 9, and z = 1.
NemiM [27]

Answer:

-20

Step-by-step explanation:

x(-y+z) [Switch in the values]

=5(-5+1) [Use distributive property and multiply]

=(-25)+5 [do the basic math]

=(-20) [final answer]

6 0
3 years ago
Read 2 more answers
(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
What is the equation of the line that passes through the point (–2,7) and has a slope of zero?
scZoUnD [109]

Answer:

y = 7

Step-by-step explanation:

Given:

x1 = -2

y1 = 7

slope (m) = 0

y - y1 = m ( x - x1 )

y - 7 = 0 ( x - (-2) )

y - 7 = 0 ( x + 2 )

y - 7 = 0

y = 7

4 0
2 years ago
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