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Svetlanka [38]
3 years ago
14

A square has an area of 150 square centimeters. Which of the

Mathematics
1 answer:
Sergio [31]3 years ago
7 0

Answer:

C) 12.2 cm

Step-by-step explanation:

To find the area of a square, you multiply length and width, which for a square, are the same. So to find the length of one of its sides, you have to take the square root of the area.

L^{2} = 150\\\\L=\sqrt{150} \\L= 12.2\\

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Nadine can send or receive a text message for $0.15 or get an unlimited number for $5.00. Write and solve an inequality to deter
34kurt

Answer:

0.15*t < 5.00

33 texts

Step-by-step explanation:

given : $0.15 per text; unlimited cost is $5.00

# of texts = t

0.15*t < 5.00

t < 33.33; since you cannot make 33.33 texts i will round down to 33 texts.

Basically you can send or receive up to 33 texts and it will be cheaper than $5.00

0.15(33) = $4.95

at 34 texts, 0.15(34) = $5.10

8 0
3 years ago
Use set builder notation to represent the following set. {... -3, -2, -1, 0} {x | x Z, x ≤ 0} {x | x N, x ≤ 0} {x | x R, x ≤ 0}
IceJOKER [234]

Answer:

\boxed{\boxed{\left\{x\ |\ x\ \epsilon\ Z,\ x \le 0}\right\}}}

Step-by-step explanation:

Set builder notation-

It is a notation for representing a set by enumerating its elements and stating the properties that its members must satisfy.

The given set is,

\left \{.... -3, -2, -1, 0 \right \}

This set is comprised of all negative integers. So symbol Z should be used in this case.

The set builder notation for this set is,

\left\{x\ |\ x\ \epsilon\ Z,\ x \le 0}\right\}

7 0
3 years ago
Daniel worked for 213 hours and earned $18.20.
OleMash [197]

Answer:

2 1/3hrs→ $18.20

1 hr→ $7.80

5 1/4hrs→ $7.80x5 1/4= $40.95

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
8 mi. 133 yd. 2 ft. – 5 mi. 107 yd. 2 ft.
Murrr4er [49]

8 mi. 133 yd. 2 ft. - 5mi. 107 yd. 2 ft.

= (8 mi. - 5 mi.) + (133 yd. - 107 yd) + (2 ft. - 2ft.)

= 3 mi. +26 yd.

<h3>= 3 mi. 26 yd.</h3>
8 0
3 years ago
Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt
Leni [432]
Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = \sqrt{Δx^2+Δy^2}
d₁ = \sqrt{(-2-1)^2+(1-4)^2}
d₂ = \sqrt{(x-1)^2+(y-4)^2}
d₁ = d₂
∴ \sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2} ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
 (x-1)²+(y-4)² = 18
from equatoin (1)  y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = \pm \sqrt{9} = \pm 3
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1)  or  (-2,7)












8 0
3 years ago
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