Is -12 greater than -5

I have just one question, could someone please answer it?? (see linked file)

Annette [7]

The solution is x=1 and y=-2

Further explanation:

Given equations are:

$\frac{7}{2}x-\frac{1}{2}y=\frac{9}{2}\\3x-y=5$

Multiplying equation no. 1 with 2 will give us:

$7x-y=9$ Eqn 3

From equation no 2:

$3x-y=5\\-y=5-3x\\y=3x-5$

Putting the value of y in eqn 3

$7x-(3x-5)=9\\7x-3x+5=9\\4x=9-5\\4x=4\\\frac{4x}{4}=\frac{4}{4}\\x=1\\Putting\ x=1\ in\ equation\ 2\\3(1)-y=5\\3-y=5\\-y=5-3\\-y=2\\y=-2\\$

The solution is x=1 and y=-2

Keywords: Linear equations, Solution set

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8 0

3 years ago

What is the difference for -22 - (-26)

Alenkasestr [34]

Answer:

4

Step-by-step explanation:

8 0

3 years ago

What are the vertex and x-intercepts of the graph of the function given below?

Travka [436]

In order to determine the vertex of this, you can complete the square. To do that, first set the equation equal to 0, then move the -35 over to the other side by adding. That gives us $x^{2} -2x=35$ . Now we can complete the square. Do this by taking half of the linear term, squaring it, and adding it in to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. So we add 1 to both sides, creating something that looks like this: $x^{2} -2x+1=35+1$ . We will do the math on the right and get 36, and the left will be expressed as the perfect square binomial we created by doing this whole process. $(x-1)^2=36$ . Now move the 36 over by subtraction and set it back to equal y and your vertex is apparent. It is (1, -36). You find the x-intercepts when y = 0. That means you need to set your original equation equal to zero and factor it. The easiest, surest way to do this is to use the quadratic formula. Doing that gives us x values of 7 and -5. And you're done!

7 0

2 years ago

Read 2 more answers

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .