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postnew [5]
4 years ago
5

Check all of the answers that are in the solution set of the inequality.

Mathematics
1 answer:
blsea [12.9K]4 years ago
7 0
Well, plug in all the values.

(sqrt)3>3

False. 3 would be more than it's square root.

(sqrt)9>3
3>3

This is false. 3 is not bigger than 3. They are equal.

(sqrt)15>3 

3.8729..... >3

That is true,

(sqrt)10>3

3.1622>3 

This true.

-3 and 0 would not work either.

So, 

"C" and "D" are correct solutions.

I hope this helps!
~kaikers


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Luke is designing a scale model of a clock tower. The design of the front of the tower is shown below.
Nat2105 [25]

Complete question :

Luke is designing a scale model of a clock tower. The design of the front of the tower is shown below. Obtain the area of the front face of the model.

Answer:

12500 mm²

Step-by-step explanation:

The front face of the mode consista of both triangle and rectangle

Area of rectangle :

Height * Base

Height = 200 m ; Base = 50

Area = 200 m * 50 m = 10,000 mm²

Area of triangle :

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1/2 * 50 * (300 - 200)

1/2 * 50 * 100

= 2500 mm²

Area of front face = Area of rectangle + Area of triangle

Area of front face = (10,000 + 2500) m²

Area of front face = 12500 mm²

3 0
3 years ago
PLEASE HELP ASAP BRAINLIEST
Y_Kistochka [10]

Area is Length x width.

Area = 4x * 5

Area = 20x

Replace A in the inequality:

Now you have 100 <= 20x <= 1,000

Now solve for x by dividing all 3 by 20:

5 <= x <=50

Check to see if they are viable:

4(5) x 5 = 20 x 5 = 100

4(50) x 5 = 200 x 5 = 1,000

Since the lowest number equals 100 and the largest number equals 1,000 all values of x would be viable. The answer is YES.

6 0
3 years ago
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3 years ago
The equation p = 1.95t² + 12.25t + 125 approximates the average sale price p of a house (in thousands of dollars) for years t si
Alborosie

Resolving   p = 1.95t² + 12.25t + 125

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We have that

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<span>is the best estimate for the price of the house in year 2021</span>
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3 years ago
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Answer to the observation is “ i noticed they’re all congruent “ and the rest are “ congruent “ for each line .
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