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3 years ago

Which system of linear inequalities Is represented by this graphed solution?

Mathematics

1 answer:

SashulF [63]3 years ago

7 0

Answer:

y ≥ 3x - 1

y < $-\frac{1}{2}x+2$

Step-by-step explanation:

Let the equation of the solid line passing through (x', y') given in graph is,

y = mx + b

Here m = slope of the line

b = y-intercept

Slope of the solid line =

m = $\frac{3}{1}$

m = 3

y - intercept 'b' = -1

Therefore, equation of the line will be,

y = 3x - 1

Since, shaded (blue) area is above the line, inequality that will represent the solution area will be,

y ≥ 3x - 1

For the dotted line,

Slope of the line (m) = $\frac{\text{Rise}}{\text{Run}}$

= $\frac{-2}{4}$

= $-\frac{1}{2}$

y-intercept (b) = 2

Equation of the dotted line → y = $-\frac{1}{2}x+2$

Since, shaded (grey color) area is below the dotted line,

Inequality representing the solution area will be,

y < $-\frac{1}{2}x+2$

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Write the equation of the parabola that has its x-intercepts at (1+<img src="https://tex.z-dn.net/?f=%5Csqrt%7B5%7D" id="TexForm

VladimirAG [237]

Answer:

$y=2x^2-4x-8$

Step-by-step explanation:

<u>Factored form of a parabola</u>

$y=a(x-p)(x-q)$

where:

p and q are the x-intercepts.
a is some constant.

Given x-intercepts:

(1+√5, 0)
(1-√5, 0)

Therefore:

$\implies y=a(x-(1+\sqrt{5}))(x-(1-\sqrt{5}))$

$\implies y=a(x-1-\sqrt{5})(x-1+\sqrt{5})$

To find a, substitute the given point (4, 8) into the equation and solve for a:

$\implies a(4-1-\sqrt{5})(4-1+\sqrt{5})=8$

$\implies a(3-\sqrt{5})(3+\sqrt{5})=8$

$\implies4a=8$

$\implies a=2$

Therefore, the equation of the parabola in factored form is:

$\implies y=2(x-1-\sqrt{5})(x-1+\sqrt{5})$

Expand so that the equation is in standard form:

$\implies y=2(x^2-x+\sqrt{5}x-x+1-\sqrt{5}-\sqrt{5}x+\sqrt{5}-5)$

$\implies y=2(x^2-x-x+\sqrt{5}x-\sqrt{5}x+\sqrt{5}-\sqrt{5}+1-5)$

$\implies y=2(x^2-2x-4)$

$\implies y=2x^2-4x-8$

6 0

1 year ago

Why is it helpful to write numbers in different ways

Goshia [24]

You should write numbers in as many ways as you possibly can to make new connections in your brain. Knowing how to write numbers in many different ways can help you solve complex problems more easily. Doing this can also reinforce the mathematical principles and logic you have memorised.

Writing one in many different ways:

1=1/1=2/2=3/3=4/4=(-1)/(-1)=(-2)/(-2)

=1.0=1.00=1.000=(1/2)+(1/2)=(1/3)+(1/3)+(1/3)

=(1/4)+(1/4)+(1/4)+(1/4)

Writing a half in many different ways:

1/2=(1/4)+(1/4)=(1/6)+(1/6)+(1/6)

=(1/8)+(1/8)+(1/8)+(1/8)=4*(1/8)

=2/4=3/6=4/8=5/10=0.5=0.50

etc...etc...

6 0

3 years ago

The figure below has a volume of 60 cubic units. How many more cubes will it take to fill the figure?

jolli1 [7]

Answer:

40 cubic units

Step-by-step explanation:

4 0

2 years ago

(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)

grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0

2 years ago

In Exercises 1-16, solve the inequality. Graph the solution<br> 1. 3y ≤ -9

murzikaleks [220]

By solving the inequality we get $y\leq -3$

What is graphing the inequality?

The act of illustrating which area of the number line contains values that will "satisfy" the specified inequality is known as graphing the inequality. Take a look at the first inequality, "x > -5." The numbers that can be used to substitute x in our inequality to produce a true statement can be shown on a graph of our inequality.

$3y\leq -9$