Tim has a 6-sided die that contains the

Jessie draws triangle ABC on a coordinate grid. The slope of line segment AB is Jessie then transforms triangle

balu736 [363]

Answer:

1) Supports Jessie's Claim

2) Does Not Support Jessi's Claim

3) Supports Jessie's Claim

4) Does Not Support Jessi's Claim

Step-by-step explanation:

The given transformations are;

1) Rotation of 180° around the origin

For a rotation of 180° around the origin, either clockwise or anti clockwise, for a given coordinate of the preimage (x, y), the coordinate of the image is (-x, -y)

Therefore, whereby the slope of the preimage, given two points (0, 0) and (2, 2), = (2 - 0)/(2 - 0) = 1

For the image with the points (0, 0) and (-2, -2), we have;

(-2 - 0)/(-2 - 0) = 1

Therefore, the slope of the preimage and the image are equal

Therefore, supports Jessie's Claim

2) For a reflection across the line y = 2, we have

We note that the line y = 2 is parallel to the x-axis

For a reflection across the x-axis, for a preimage (x, y), we have the coordinates of the image (x, -y)

However for the reflection across the line y = 2, we have;

For a preimage, (x, y), the coordinate of the image is (x, -y+4)

Given two points, of the preimage (0, 0) and (2, 2), we have the image given as (0, 4) and (2, -2 + 4) = (2, 2);

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (2 - 4)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

3) For a translation up 1.25 units, we note that the difference in the y and x values of the coordinates of the preimage and the image will be equal when finding the slope, and therefore, the slope of the figure of the preimage and the slope of the figure of the image will be equal

Therefore, supports Jessie's Claim

4) For a reflection across the x-axis, a point on the preimage, with coordinates (x, y) will form a point on the image with coordinates (x, - y)

For a preimage with points (0, 0) and (2, 2), we have the image as (0, 0) and (2, -2)

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (-2 - 0)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

6 0

3 years ago

Which expression can be used to find the price of a $400 telescope after a 32% markup? Select all that apply.

elena-s [515]

<h3>There are 2 answers: choice C, choice D</h3>

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Explanation:

32% = 32/100 = 0.32

32% of 400 = 0.32*400 = 400(0.32), which adds onto the original 400 to get 400+400(0.32). This is why choice D is one of the answers.

We can factor out the GCF 400 to go from 400+400(0.32) to 400(1+0.32) which then simplifies to 400(1.32) or just 400*1.32. This shows choice C is the other answer. Using a calculator,

400+400(0.32) = 400 + 128 = 528

400*1.32 = 528

meaning that 400+400(0.32) = 400*1.32

The other answer choices result in other values, showing that they aren't equivalent to 528.

6 0

3 years ago

A cup of coffee contains about 100 mg of caffeine. Caffeine is metabolized and leaves the body at a continuous rate of about 12%

Mamont248 [21]

Answer:

a. $A = C_{0}(1-x)^t\\x: percentage\ of \ caffeine\ metabolized\\$

b. $\frac{dA}{dt}= -11.25 \frac{mg}{h}$

Step-by-step explanation:

First, we need tot find a general expression for the amount of caffeine remaining in the body after certain time. As the problem states that every hour x percent of caffeine leaves the body, we must substract that percentage from the initial quantity of caffeine, by each hour passing. That expression would be:

$A= C_{0}(1-x)^t\\t: time \ in \ hours\\x: percentage \ of \ caffeine\ metabolized\\$

Then, to find the amount of caffeine metabolized per hour, we need to differentiate the previous equation. Following the differentiation rules we get:

$\frac{dA}{dt} =C_{0}(1-x)^t \ln (1-x)\\\frac{dA}{dt} =100*0.88\ln(0.88)\\\frac{dA}{dt} =-11.25 \frac{mg}{h}$