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1 year ago

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PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

sformed under the rule (x,y) (x-1,y-2) then does T' form a right angle at <GRT'?

1 answer:

Anna007 [38]1 year ago

6 0

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

$m=\frac{y_2-y_1}{x_2-x_1}$

The product of slopes of two perpendicular lines is -1.

Slope of GR is

$\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}$

Slope of RT is

$\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1$

Product of slopes of GR and RT is

$\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1$

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

$(x,y)\rightarrow(x-1,y-2)$

Then the coordinates of T' are

$(2,6)\rightarrow(2-1,6-2)$

$(2,6)\rightarrow(1,4)$

Slope of RT' is

$\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}$

Product of slopes of GR and RT' is

$\frac{-4}{3}\times \frac{3}{4}=-1$

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

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We have been given an equation $ae^{ct}=d$ . We are asked to solve the equation for t.

First of all, we will divide both sides of equation by a.

$\frac{ae^{ct}}{a}=\frac{d}{a}$

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Now we will take natural log on both sides.

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Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

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Now we will divide both sides by c as:

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2 years ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

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Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

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To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

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$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

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$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .

Apparently $\displaystyle c = \frac{16}{10} = \frac{8}{5}$ works. $c = 1.6$ while $\displaystyle \frac{1}{c} = \frac{5}{8} = 0.625$ .

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$Helen’s\ house\ is\ located\ on\ a\ rectangular\ lot\ that\ is\ \frac{9}{8}\ miles\ by\ \frac{9}{10}\ mile.$

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