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3 years ago

Solve the inequality 7+7(x+4)>21

Mathematics

1 answer:

Kryger [21]3 years ago

8 0

Answer:

x>-2

Step-by-step explanation:

Hi there!

We are given the inequality 7+7(x+4)>21, and we want to solve it

Solving an inequality is a lot like solving an equation; we want to isolate x on one side by itself.

So to start, we'll preform the distributive property on 7(x+4) (multiply x by 7, then multiply 4 by 7 and add those two together).

7+7x+28>21

Now combine like terms.

7x+35>21

Subtract 35 from both sides

7x>-14

Divide both sides by 7

x>-2

Hope this helps!

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Solve the following equation.-3 = x/5

Umnica [9.8K]

I hope this helps you

x/5= -3

x= -3.5

x= -15

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3 years ago

Find the perimeter of the triangle. Recall that the perimeter of a figure is the distance around a figure.

Svet_ta [14]

Answer:

1 inch

Step-by-step explanation:

The perimeter is the sum of the lengths of the sides:

Perimeter = 3/23 + 9/23 + 11/23 = (3 + 9 + 11)/23 = 23/23 = 1

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2 years ago

Please help! Correct answer only!

Tcecarenko [31]

Answer:

" Expected Payoff " ⇒ $ 1.56 ; Type in 1.56

Step-by-step explanation:

Consider the steps below;

$Tickets That Can Be Entered - 1 Ticket,\\Total Tickets Entered - 1000 Tickets,\\\\Proportion - 1 / 1000,\\Money One ( First Ticket ) = 820 Dollars,\\Money One ( Second Ticket ) = 740 Dollars,\\\\Proportionality ( First ) - 1 / 1000 = x / 820,\\Proportionality ( Second ) - 1 / 1000 = x / 740\\\\1 / 1000 = x / 820,\\1000 * x = 820,\\x = 820 / 1000,\\x = 0.82,\\\\1 / 1000 = x / 740,\\1000 * x = 740,\\x = 740 / 1000,\\x = 0.74\\\\Conclusion ; " Expected Payoff " = 0.82 + 0.74 = 1.56$

Solution; " Expected Payoff " ⇒ $ 1.56

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3 years ago

Brenda throws a dart at this square-shaped target:

Klio2033 [76]

Answer: See below

Step-by-step explanation:

Area of the square = l*w = 11*11 = 121

Area of the circle = pi r^2 = pi*1^2 = 3.142

Part A

P(hitting the circle) = 3.142 / 121 = 0.026 which is closer to 0.

Part B

P(hitting the outside portion) = 1 - 0.026 = 0.974 which is closer to 1.

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avanturin [10]

Congruent
SSS (because of Pythagoras)

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