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1 year ago

Which number line represents the solution set for the inequality 3(8 – 4x) < 6(x – 5)?

1 answer:

5 0

The solution will be a number line from negative 5 to 5 in increments of 1. An open circle is at 3 and a bold line starts at 3 and is pointing to the right.

<h3>Inequality expression</h3>

Inequalities are equations that are separated by an equal sign. Given the inequality expression

3(8 – 4x) < 6(x – 5)

Expand

3(8) - 3(4x) < 6x - 6(5)

24 - 12x < 6x - 30

Collect the like terms

-12x - 6x < -30 - 24

-18x < -54

x > 3

Hence the solution will be a number line from negative 5 to 5 in increments of 1. An open circle is at 3 and a bold line starts at 3 and is pointing to the right.

Learn more on inequality here: brainly.com/question/11613554

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In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.

bezimeni [28]

Given:

In the circle, $m(\overarc{VUX})=152^\circ$ and $m(\angle MUV)=77^\circ$ .

To find:

The following measures:

(a) $m\angle VUX$

(b) $m\angle UXW$

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

$m(VUX)=2\times m\angle VWX$

$152^\circ=2\times m\angle VWX$

$\dfrac{152^\circ}{2}=m\angle VWX$

$76^\circ=m\angle VWX$

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

$m\angle VUX+m\angle VWX=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle VUX+76^\circ=180^\circ$

$m\angle VUX=180^\circ-76^\circ$

$m\angle VUX=104^\circ$

Now,

$m\angle UXW+m\angle UVW=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle UXW+77^\circ =180^\circ$

$m\angle UXW=180^\circ-77^\circ$

$m\angle UXW=103^\circ$

Therefore, $m\angle VUX=104^\circ$ and $m\angle UXW=103^\circ$ .

7 0

2 years ago

How do you rewrite 6/9 in two different ways

Naya [18.7K]

We have to represent the fraction $\frac{6}{9}$ in two different ways.

Let us multiply the numerator and denominator of the given fraction by '2'.

Therefore, $\frac{6}{9} = \frac{6 \times 2}{9 \times 2} = \frac{12}{18}$

Therefore, $\frac{12}{18}$ is the first way to represent the given fraction.

Now, Let us multiply the numerator and denominator of the given fraction by '3'.

Therefore, $\frac{6}{9} = \frac{6 \times 3}{9 \times 3} = \frac{18}{27}$

Therefore, $\frac{18}{27}$ is the second way to represent the given fraction.

Therefore, $\frac{12}{18}$ and $\frac{18}{27}$ are the different ways to represent the fraction $\frac{6}{9}$ .

4 0

2 years ago

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

3 years ago

Jenna bought two more pairs of jeans. If both of them are blue, what percentage of her jeans is blue now? If both of them are no

Sergeeva-Olga [200]

What you must keep in mind is that there are 4 jeans in total, which represents 100%
When she bought two jeans, we have the following information:
If both of them are blue:
50% of their jeans is blue now, that is:
x = (50/100) * 4 = 2
2 jeans are blue.
This means that she did NOT have 2 blue jeans before.
Thus:
If they are not blue:
We have 4 jeans that are not blue.
Then, 0% of her jeans is blue now
answer:
0% percentage of her jeans is blue now

7 0

3 years ago

The diameter of a Frisbee is 12 inches. what's the area of the Frisbee?

xenn [34]

A= πr² = π × 6² = "113.1" is the area of the frisbee.

Please put me as brainliest.

7 0

3 years ago