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Paha777 [63]
3 years ago
5

Which is true about the solution to the system of inequalities shown? y < x – 1 y < x – 3 All values that satisfy y < x

– 1 are solutions. All values that satisfy y < x – 3 are solutions. All values that satisfy either y < x – 1 or y < x – 3 are solutions. There are no solutions.

Mathematics
2 answers:
Masteriza [31]3 years ago
7 0
<span>solution of a system of linear equations</span>
Naily [24]3 years ago
5 0

Answer:

Option (2) is correct.

All values that satisfy y < x – 3 are solutions.

Step-by-step explanation:

Given two inequality y < x – 1  and y < x – 3

We have to choose out of given options that is true  about the solution to the system of inequalities given.

Step 1: We first plot both the inequality in the graph.(as shown below)

Step 2: Choose a test point to determine which side of the line needs to be shaded.

Step 3: the solution to the system will be the area where the shadings from both the inequality overlap one another.

Thus, All values that satisfy y < x – 3 are solutions.

Thus, Option (2) is correct.

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3 years ago
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What is 3/5 2/3 and 0.65 from least to greatest
Anna35 [415]
3/5 = 9/15 = 36/60

2/3 = 10/15 = 40/60

0.65 = 65/100 = 13/20 = 39/60

Therefor, the answer is 3/5, 0.65, 2/3.





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3 years ago
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How do I do this? PR bisects angle SPT. Find TR
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3 years ago
2x-1<br> Hahahahahhaahhahahahahahahhahaha
sweet-ann [11.9K]

Answer:

x=1/2

Step-by-step explanation:

Set up the equation: 2x-1=0

Add one to both sides. 2x=1

Divide two by both sides: x=1/2

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4 0
3 years ago
Read 2 more answers
The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.
Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m  + 40 m  + c = 120 m

61 m +  c = 120 m

c = 120 m - 61 m

c = 59 m

                         B

                     ↓            ↓  

                  ↓                          ↓

                ↓                                       ↓

            A →  →  → →  →  → →  → →    →    →  C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0

Angle B is calculated as follows;

Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0

Angle C is calculated as follows;

Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0

Therefore, the largest angle of the field is 149⁰.

       

8 0
3 years ago
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