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2 years ago

find the common difference and the three terms in the sequence after the last one given -39 -33 -27 -21

Mathematics

1 answer:

Serggg [28]2 years ago

6 0

Answer:

Common difference:6 3 terms: -15, -9. -3

Step-by-step explanation:

-39, -33, -27, and -21 all have a difference of 6.

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Half of 836 - triple of 87

barxatty [35]

418-261=157
Answer: 157
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2 years ago

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I don't understand this question at all please help if you do

lorasvet [3.4K]

Answer:

Answer is $\frac{4}{3}$

Step-by-step explanation:

To find the interval of x. Use our equations to equal each other.

$-x^2+2x=y,y=0$

$-x^2+2x=0\\-x(x-2)=0\\-x=0 , (x-2)=0\\x=0 ,2$

$\int\limits^2_0 {-x^2+2x} \, dx$

Integrate.

$\frac{-x^3}{3}+x^2\\(\frac{-2^3}{3}+2^2)-[\frac{-0^3}{3}+0^2]\\-\frac{8}{3} +4-0\\-\frac{8}{3}+\frac{12}{3} =4/3$

Using Desmos I have Graphs of both of the equations you have provided. The problem asks us to find the shaded region between those curves/equations.

Proof Check your interval of x.

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2 years ago

Find the standard form of the equation of the parabola with the given characteristics. Vertex: (-6, 1); focus: (-6, 0)

GREYUIT [131]

Answer: $(x+6)^2=-4(y-1)$

Step-by-step explanation:

Given

Vertex of the parabola $(-6,1)$

Focus of the parabola $(-6,0)$

As the x coordinate of vertex and focus is same and focus lie below the vertex, therefore the parabola is the type of

$(x+6)^2=-4a(y-1)$

distance between Vertex and focus is 1 unit

$\therefore a=1$

Parabola becomes

$\Rightarrow (x+6)^2=-4(y-1)$

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3 years ago

Graph a triangle (STU) and reflect it over the y-axis to create triangle ST'U'.

lukranit [14]

The x-coordinates of $\triangle S'T'U'$ will be the negation of the x-coordinates of $\triangle STU$

The line segment from S to the y-axis equals the line segment from S' to the y-axis. Similarly, the line segment from T to the y-axis equals the line segment from T' to the y-axis

See attachment for $\triangle STU$ and $\triangle S'T'U'$

In order to solve this question, I will make the following assumptions.

Assume that the coordinates of $\triangle STU$ are

$S = (4,5)$

$T = (5,9)$

$U=(3,8)$

Refer to attachment for illustrations

(1) Reflect $\triangle STU$ over y-axis and describe the transformation

To reflect $\triangle STU$ across the y-axis, the following rule must be followed

$(x,y) \to (-x,y)$

This means that:

$S = (4,5) \to S' = (-4,5)$

$T = (5,9) \to T' = (-5,9)$

$U=(3,8) \to U'=(-3,8)$

The description of the transformation is as follows:

Notice that the signs of the x-coordinates $\triangle STU$ and $\triangle S'T'U'$ of both triangles are different.

In other words, if the x-coordinate of one is positive, then the other will have a negative x-coordinate; and vice versa.

(2) Compare the segments and the line of reflection

To reflect across the y-axis means that the reflecting line is the y-axis, itself.

The distance between a point to the y-axis is the absolute value of the x-coordinate.

So, the distance between S and the y-axis is:

$S = |4| = 4$

The distance between S' and the y-axis is:

$S' = |-4| = 4$

We can conclude that the two line segments are equal.

This is the same for other point T and T' because of the formula used above.

From T and T' to the y-axis is:

$T =|5| =5$

$T' =|-5| =5$

find the common difference and the three terms in the sequence after the last one given -39 -33 -27 -21​

find the common difference and the three terms in the sequence after the last one given -39 -33 -27 -21