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3 years ago

Which image that would portray the rotation of the given preimage and the rotation arrow.

Mathematics

2 answers:

Karo-lina-s [1.5K]3 years ago

5 0

Answer:

First figure as shown below is the correct option.

Step-by-step explanation:

We are given that,

The pre-image is rotated in the direction of the arrow to form an image.

It is required to find the image formed.

Since, the arrow is rotating counter-clockwise, we get that,

The star in the pre-image will rotate upwards towards the left.

So, according to the options, we have,

The first image where the star is at the highest height will be the image after rotating the pre-image.

Hence, the first figure as shown below is the correct option.

Brut [27]3 years ago

4 0

Is the arrow the length of the rotation? 90 degrees? if so, the second picture is, with the star at the highest point of the other pictures.

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Which polynomial matches the expression -x(x – 5)?x2 – 5x x2 + 5x -x2 + 5x -2x – 5

Leno4ka [110]

Answer:

-x2 + 5x

Step-by-step explanation:

See below.

The graphs are the same

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3 years ago

Write a story problem that can be solved by finding the sum of 506,211 and 424,809 then solve the preoblem

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3 years ago

How many solution does this have and how do you solve it

Vadim26 [7]

Answer:

Infinite solutions

Step-by-step explanation:

To determine the number of solutions a system has, you need its slope. If the slopes are the same, check if the equations are equivalent or not.

To find the slope, rearrange the equation to isolate "y", which makes it slope-intercept form y = mx + b.

"x" and "y" represent points on the line.

"m" represents the slope, which is how steep a line is.

"b" represents the y-intercept (where the graph hits the y-axis).

Isolate "y" in both equations:

x + 3y = 0

3y = -x Subtract 'x' from both sides

y = -x/3 Divide both sides by 3.

y = $-\frac{1}{3}x$

The slope is -1/3.

9y = -3x

y = -3x/9 Divide both sides by 9

y = $-\frac{3}{9}x$ Reduce the fraction to lowest terms

y = $-\frac{1}{3}x$

The slope is -1/3, but the two equations are the same.

Since the equations are the same, or equivalent, the two lines are always touching. The solution of a system means where two lines intersect, or meet. Therefore, there are infinite solutions.

5 0

3 years ago

Consider the parabola r(t)equalsleft angle at squared plus 1 comma t right angle, for minusinfinityless thantless thaninfinity

kodGreya [7K]

Given:- $r(t)=< at^2+1,t>$ ; $-\infty < t< \infty$ , where a is any positive real number.

Consider the helix parabolic equation :

$r(t)=< at^2+1,t>$

now, take the derivatives we get;

$r{}'(t)=$

As, we know that two vectors are orthogonal if their dot product is zero.

Here, $r(t) and r{}'(t)$ are orthogonal i.e, $r\cdot r{}'=0$

Therefore, we have ,

$< at^2+1,t>\cdot < 2at,1>=0$

$< at^2+1,t>\cdot < 2at,1>=$

$=2a^2t^3+2at+t$

$2a^2t^3+2at+t=0$

take t common in above equation we get,

$t\cdot \left (2a^2t^2+2a+1\right )=0$

⇒ $t=0$ or $2a^2t^2+2a+1=0$

To find the solution for t;

take $2a^2t^2+2a+1=0$

The number $D = b^2 -4ac$ determined from the coefficients of the equation $ax^2 + bx + c = 0.$

The determinant $D=0-4(2a^2)(2a+1)$ $=-8a^2\cdot(2a+1)$

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola $r(t)=< at^2+1,t>$ i.e <1,0>

6 0

3 years ago

Please help I don't get this one

Semenov [28]

Its A you just have to count the lines and say the numbers 1-4

7 0

3 years ago