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3 years ago

Solve the system of inequalities: a) .4x−1≤0 2.3x≥4.6

Mathematics

1 answer:

OleMash [197]3 years ago

4 0

Answer:

(a) $x \leq 2.5$

(b) $x \geq 2$

Step-by-step explanation:

Solving (a):

$0.4x - 1 \leq 0$

Add 1 to both sides

$0.4x - 1 + 1\leq 0 + 1$

$0.4x \leq 1$

Solve for x

$x \leq 1/0.4$

$x \leq 2.5$

Solving (b):

$2.3x \geq 4.6$

Solve for x

$x \geq 4.6/2.3$

$x \geq 2$

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3 years ago

Find the value of x. Round to the nearest tenth. Answers A-12.5 B-10 C-13 D-9.7!

kvasek [131]

Answer: A. 12.5

Step-by-step explanation:

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3 years ago

Read 2 more answers

Distance between (-3,1) (1,5)

Mamont248 [21]

Answer & Step-by-step explanation:

For this problem, we will use the distance formula.

$d=\sqrt{(x2-x1)^2+(y2-y1)^2}$

For our (x1, y1), we will use (-3, 1). For our (x2, y2), we will use (1, 5). Now let's plug in our numbers and solve the problem.

$d=\sqrt{(1-(-3))^2+(5-1)^2}$

$d=\sqrt{4^2+4^2}$

$d=\sqrt{16 + 16}$

$d=\sqrt{32}$

$d=5.657$

So, the distance between (-3,1) and (1,5) is 5.657 units.

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2 years ago

Please help me with this explain!!!!!

erik [133]

Given:
Cylinder: radius = 8 yd; height = 4 yd
Surface Area = 2 π r h + 2 π r²
SA = 2 * 3.14 * 8 yd * 4yd + 2 * 3.14 * (8yd)²
SA = 200.96 yd² + 401.92 yd²
SA = 602.88 yd²
Volume = π r² h
V = 3.14 * (8yd)² * 4yd
V = 803.84 yd³

Dimension is cut in half. radius = 4yds ; height = 2yds
S.A = 2 * 3.14 * 4yd * 2yd + 2 * 3.14 * (4yd)²
S.A = 50.24 yd² + 100.48 yd²
SA = 150.72 yd²
V = 3.14 * (4yd)² * 2yd
V = 100.48 yd³

SA = 602.88 yd² - 150.72yd² = 452.16 yd²
V = 803.84 yd³ - 100.48 yd³ = 703.36 yd³

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3 years ago

Aria studied the traffic trends in India. She found that the number of cars on the road increases by 10% each year. If there wer

denpristay [2]

Answer:

8,800,000cars

Step-by-step explanation:

Given the initial amount of cars = 80,000,000

If there is 10 % increase in the first year;

Increment = 10% of 80,000,000

Increment = 0.1 * 80,000,000

Increment = 8,000,000

Amount of car in year 2 = 80million + 8million = 88million

Year3;

Amount of car in year 3 = 88,000,000+(10%of88million)

Amount of car in year 3 = 88,000,000+8,800,000

Amount of car in year 3 =96,800,000

Differences in the number of cars = 96,800,000 - 88,000,000

Differences in the number of cars = 8,800,000cars

Hence there are 8,800,000cars in years 3 than year 2

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3 years ago