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3 years ago

What is the absolute value of |-15|? Explain!

Mathematics

2 answers:

GrogVix [38]3 years ago

7 0

Answer: 15

Step-by-step explanation: anything in those two straight lines is automatically positive

EleoNora [17]3 years ago

6 0

Answer:

Step-by-step explanation:

The absolute value is how far away it is from 0. If it's -15, it is 15 spaces away from 0. If it is 15, it is still 15 spaces away from 0. The bars around -15 are the absolute value bars.

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Every square is a rectangle.

Hoochie [10]

<span>Every square is a rectangle. Always

</span>A<span> rectangle is any four-sided with four right angles and opposite sides are equal.
Squares has four right angles, so square is a special type of rectangle with all sides equal.
</span>

7 0

2 years ago

Pinto is planning to paint the walls of a bedroom that is 20 feet long, 15 feet wide, and 8 feet high. Is he too has 1 gallon of

eimsori [14]

Answer:

0.4 gallons

Step-by-step explanation:

Length of a bedroom= 20 feet

Breadth of a bedroom= 15 feet

Height of a bedroom= 8 feet

Area of walls of a bedroom= 2(l+b)×h

Area of walls of a bedroom=2(20+15)×8 $ft^{2}$

Area of walls of a bedroom= 2× 35×8 $ft^{2}$

Area of walls of a bedroom=560 $ft^{2}$

Amount of paint in 1 gallon = 400 $ft^{2}$

Number of gallons required= $\frac{560}{400}$

Number of gallons required= 1.4 gallons

Hence, 0.4 extra gallons required

5 0

3 years ago

PLEASE HELP ME IM BEGGING YALL

drek231 [11]

Answer:

False. Correct ratio is $\frac{12}{37}$

Step-by-step explanation:

the ratio of sine is opposite over hypotenuse

(opposite is the side across from the angle and hypotenuse it the slant line in the triangle)

7 0

2 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$