Which statement is the correct interpretation of the inequality –2 > –8?
On a number line, −2 is located to the right of –8.
On a number line, −2 is located to the left of –8.
On a number line, −2 is located to the left of 0 and –8 is located to the right of 0.
On a number line, −2 is located to the right of 0 and –8 is located to the left of 0.

Answer: The answer is C.

A Linear parent function is a function of the form f(x) = x

or y = x. This is the simplest of the linear functions, where the slope is 1 and the function pases for the point (0,0) (which is the same of saying that the intercept is 0).

a) isnt a linear function.

b) is a quadratic function.

d) is also a linear function, but it has a slope = 2, so it isn't the linear parent function.

Answer:

has two rays that share a common endpoint

Step-by-step explanation:

An angle is formed by a two rays that share a common endpoint.  The distance between the two rays is what makes the measure of the angle wider or smaller.  Each ray extends in one direction forever; however since they share an endpoint, the rays do not extend in both directions forever.  There is only one endpoint, not two, and there is only one center point (vertex), not two.

V=a^3 and volume=Length × Width × Height are the two ways to calculate the volume of cube.

Three-dimensional space is a geometric setting in which three values  are required to determine the position of an element.

A cube is a three-dimensional figure with 6 faces, 8 vertices and 12 edges.

In a cube all the sides are equal like a square.

So the volume of cube can be measured by

V=a^3

where a is the measure of a side.

And the volume of cube can also be found by

volume=Length × Width × Height

Hence V=a^3 and volume=Length × Width × Height are the two ways to calculate the volume of cube.

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The whole container can hold [tex]5\frac{5}{6}[/tex] liters, calculated by setting up a proportion based on the part of the container currently filled with water ([tex]2\frac{1}{3}[/tex]  liters representing 2/5 of the container).

To determine the full capacity of the container, we set up a proportion. Since [tex]2\frac{1}{3}[/tex]  liters represent 2/5 of the container's capacity, we want to find out how much 5/5 (the whole) of the container can hold. To set up the proportion, we have:

(2/5) of the container = [tex]2\frac{1}{3}[/tex]  liters
(5/5) of the container = x liters
Then, we multiply [tex]2\frac{1}{3}[/tex]  liters by 5/2 to solve for x:

x = ([tex]2\frac{1}{3}[/tex] liters) * (5/2)

x = (7/3 liters) * (5/2)

x = 35/6 liters

x =  [tex]5\frac{5}{6}[/tex] liters

So the whole container would hold  [tex]5\frac{5}{6}[/tex] liters.

Answer:

1,890 is the sum of the first 30 terms of the equation .

Step-by-step explanation:

[tex]a_n=4n+1[/tex]

if , n = 1

[tex]a_1=4\times 1+1=5[/tex]

If, n = 2

[tex]a_2=4\times 2+1=9[/tex]

If, n = 3

[tex]a_3=4\times 3+1=13[/tex]

The arithmetic sequence comes out to be:

5, 9, 13........

a = 5

d = [tex]a_2=a_1=9-5=4[/tex]

The sum of first 30 terms is given by:

[tex]S_n=\frac{n}{2}\times (2a+(n-1)d)[/tex]

[tex]S_{30}=\frac{30}{2}\times (2\times 5+(30-1)\times 4)[/tex]

[tex]S_{30}=1,890[/tex]

1,890 is the sum of the first 30 terms of the equation .

Answer: Conditional statement

Step-by-step explanation:

In discrete mathematics, a conditional statement denoted by "if R then S " is an if-then statement where R represents a hypothesis and S represents a conclusion.

For example : If a number is divisible by 6 ,then it must be divisible by 2 ."

Here, Hypothesis R: " number is divisible by 6"

Conclusion S: "it must be divisible by 2"

Hence, an conditional statement  has the form " if A, then B" .

Answer: 2(x+3)(x+3)

i just took the test

a square needs to rotate 90 to coincide with  itself

 but the hexagon would need to rotate 180

the answer is 180 degrees

Answer:

Option B 180° is the right answer.

Step-by-step explanation:

Given is : GHIJ is a square and ABCDEF is a regular hexagon having common centre O.

We know that a square is symmetric at a rotational angle of 90°.

A regular hexagon is symmetric at a rotational angle of 60°.

We will find the LCM of 90 and 60, to know how much to rotate, to get a preimage.

Moreover, a hexagon needs to rotate 180 degrees to coincide with itself.

LCM(60°,90°)=180°.

So, the correct option is B. 180°.

To graph the equation y = 4/5x - 7, plot the y-intercept at -7 and use the slope to find a second point before drawing a line through the two points.

To graph the equation y = 4/5x - 7, follow these steps:

Answer:

[tex]x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}[/tex] shows correct substitution of the values a, b, and c  from the given quadratic equation  [tex]-3x^2-2x+6=0[/tex]  into quadratic formula.

Step-by-step explanation:

Given: The quadratic equation [tex]-3x^2-2x+6=0[/tex]

We have to show the correct substitution of the values a, b, and c from the given quadratic equation  [tex]-3x^2-2x+6=0[/tex]  into quadratic formula.

The standard form of quadratic equation is [tex]ax^2+bx+c=0[/tex] then the solution of quadratic equation using quadratic formula is given as [tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

Consider the given  quadratic equation [tex]-3x^2-2x+6=0[/tex]

Comparing with general  quadratic equation, we have

a = -3 , b = -2 , c = 6

Substitute in quadratic formula, we get,

[tex]x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}[/tex]

Simplify, we have,

[tex]x_{1,2}=\frac{2\pm\sqrt{76} }{-6}[/tex]

Thus, [tex]x_{1}=\frac{2+\sqrt{76} }{-6}[/tex] and x_{2}=\frac{2-\sqrt{76} }{-6}

Simplify, we get,

[tex]x_1=-\frac{1+\sqrt{19}}{3},\:x_2=\frac{\sqrt{19}-1}{3}[/tex]

Thus, [tex]x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}[/tex] shows correct substitution of the values a, b, and c  from the given quadratic equation  [tex]-3x^2-2x+6=0[/tex]  into quadratic formula.

You can use the standard form of quadratic equation to find the correct substitution of the values a, b and c.

The correct substitution of the value a,b and c from the given quadratic equations into the solution form is [tex]x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}[/tex]

The standard form of quadratic equations is [tex]ax^2 + bx + c = 0[/tex]

[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

From given equation [tex]0 = -3x^2 -2x + 6[/tex], we deduce that a = -3, b = -2 and c = 6.

Thus, substituting these values in the solutions' equation:

[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ \\ x = \dfrac{2 \pm \sqrt{4 + 72}}{-6} = \dfrac{2\pm 2\sqrt{19}}{-6} = \dfrac{-1 \mp 2\sqrt{19}}{3}[/tex]

Thus, the correct substitution of the value a,b and c from the given quadratic equations into the solution form is [tex]x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}[/tex]

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Answer:

one of these is wrong cause i got a three out of four but thats better than nothing i guess.

Step-by-step explanation:

x≤0, y=1, x≥1, this one is the one i think is wrong but y=-1

Answer:

40 x 50 = 2,000

Step-by-step explanation:

When we round a number, first identify the digit to round. Then proceed with the following two rules:

(a) If the number to the right of the digit to round is 0, 1, 2, 3, or 4 (that is, less than 5), then keep the digit to round and change the rest of the digits to the right of the digit to round to zero.

(b) If the number to the right of the digit to be rounded is 5, 6, 7, 8, or 9 (that is, greater than or equal to 5), then add one to the digit to round and change the rest of the digits to the right of the digit to round to zero.

In our case, we apply the two rules, rounding to the nearest ten, as follows:

For 38 = 40

Apply rule (b) since 8 is greater than or equal to 5, then add 1 to 3 to get 4, then change 8 to zero. The result is 40.

For 47 = 50

Apply rule (b) since 7 is greater than or equal to 5, then add 1 to 4 to get 5, then change 7 to zero. The result is 50.

Now, multiply to estimate the amount of grapes that Jerry customers buy every day.

40 x 50 = 2,000

Hope this helps!

divide total weight by weight of each box

10250 / 20.5 = 500 boxes total

Answer: $25

Step-by-step explanation: you are not trying to make a profit. You are just trying to cover your cost and they cost $25 to make.

The dashed line segments form 30-degree angles in the given diagram of a regular hexagon and triangle.

In the given diagram, hexagon GIKMPR and triangle FJN are regular. The dashed line segments form 30-degree angles.

Since hexagon GIKMPR is regular, all its interior angles are equal. Each angle of a regular hexagon is 120 degrees, so angle GIK is 120 degrees.

The angle at GIK can be split into two equal angles by the dashed line segment, so each of these angles measures 60 degrees.

Similarly, since triangle FJN is regular, each of its angles measures 60 degrees.

Therefore, the dashed line segments form two 30-degree angles each side of the regular hexagon and triangle.

The correct answer is option d. The image of [tex]\overline{OQ}[/tex] after a rotation of 300 degrees about point O is [tex]\overline{OF}[/tex].

To find the image of line segment [tex]\( \overline{OQ} \)[/tex] after a rotation of [tex]\( 300^\circ \)[/tex] about point O, we need to understand that a rotation of [tex]\( 300^\circ \)[/tex] is equivalent to rotating [tex]\( 60^\circ \)[/tex] clockwise. This is because rotating [tex]\( 360^\circ \)[/tex] would bring us back to the original position, so a rotation of [tex]\( 300^\circ \)[/tex] is the same as rotating [tex]\( 360^\circ - 60^\circ = 300^\circ \)[/tex].

Given that line segment [tex]\( \overline{OQ} \)[/tex] forms a [tex]\( 30^\circ \)[/tex] angle with the dashed lines, after a rotation of [tex]\( 300^\circ \)[/tex], it will form a [tex]\( 30^\circ \)[/tex] angle with the new position of the dashed lines.

Therefore, the image of line segment [tex]\( \overline{OQ} \)[/tex] after the rotation is [tex]\( \overline{OF} \)[/tex].

The complete question

The hexagon GIKMPR and FJN are regular. The dashed line segments form 30 degree angles.

Find the image of [tex]\overline{OQ}[/tex] after a rotation of 300 degree about point O.

a. [tex]\overline{OJ}[/tex]

b. [tex]\overline{OH}[/tex]

c. [tex]\overline{OQ}[/tex]

d. [tex]\overline{OF}[/tex]