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NNADVOKAT [17]
3 years ago
7

Solve for x. −12x+13>35 Drag and drop a number or symbol into each box to correctly complete the solution.

Mathematics
1 answer:
Katen [24]3 years ago
7 0

Answer:

x < -11/6.

Step-by-step explanation:

−12x + 13 > 35

-12x > 35 - 13

-12x > 22

Divide both sides by -12 and invert the inequality sign:

x < -22/12

x < -11/6.

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What is the area of a circle with a diameter of 12.6 in.?
Zepler [3.9K]

Hey there! I'm happy to help!

To find the area of a circle, you square the radius and then multiply by pi (3.14 in our case).

The radius is half of the diameter.

12.6/2=6.3

We square this.

6.3²=39.69

We multiply by 3.14

39.69×3.14=124.6266

We round to the nearest hundredth, giving us an area of 124.63 in².

Now you can find the area of a circle! Have a wonderful day! :D

5 0
4 years ago
Y=5x-4
anyanavicka [17]

Answer:

\displaystyle -5x + y = -92\:or\:y = 5x - 92

Step-by-step explanation:

−12 = 5[16] + b

80

\displaystyle -92 = b \\ \\ y = 5x - 92

If you want it in <em>Standard</em><em> </em><em>Form</em>:

y = 5x - 92

- 5x - 5x

__________

\displaystyle -5x + y = -92

I am joyous to assist you anytime.

7 0
3 years ago
What are the 6 trigonometry functions
EastWind [94]

Sine, Cosine, Tangent, Cosecant (opposite of Sine), Secant (opposite of Cosine), and Cotangent (opposite of Tangent)

3 0
4 years ago
What is the third quartile of this data set 21,24,25,28,2935,37,39,42
valentinak56 [21]

Answer:

Definitions:

The third quartile, denoted by Q3 , is the median of the upper half of the data set. This means that about 75% of the numbers in the data set lie below Q3 and about 25% lie above Q3;

The upper half of a data set is the set of all values that are to the right of the median value when the data has been put into increasing order;

First, we verify data in increasing order:

21, 24, 25, 28, 29, 35, 37, 38, 42 ( okay);

The median is 29;

Therefore, the upper half of the data is {35, 37, 39, 42};

Q3 = ( 37 + 39 ) / 2 = 76 / 2 = 38;

Step-by-step explanation:


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4 years ago
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ycow [4]
1 \text{ u} \cdot 1 \text{ u}=1 \text{ u}^2
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