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$Name: qquad Hadley crisman Date: qquad Per: 7 th 7 th Unit 7: PC Homew This is a 2-page documer Directions: If each quadrilateral below is a parallelogram, 1. {:[MN=],[KN=],[m/_K=],[m/_L=],[m/_M=]:} Given PQ=24,PS=19,PR=42,TQ=10,m/_PQR=10$

Name: $\qquad$ Hadley crisman $\newline$ Date: $\qquad$ Per: $\newline$ $7$ th $\newline$ $7$ th $\newline$ Unit $7$ : PC $\newline$ Homew $\newline$ This is a $2$ -page documer $\newline$ Directions: If each quadrilateral below is a parallelogram, $\newline$ $1$ . $\newline$ $\begin{array}{l} M N= \\ K N= \\ m \angle K= \\ m \angle L= \\ m \angle M= \end{array}$ $\newline$ $3$ . Given $P Q=24, P S=19, P R=42, T Q=10, m \angle P Q R=10$

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Compare linear and exponential growth

Full solution

Q. Name:

\qquad

Hadley crisman

\newline

Date:

\qquad

Per:

\newline

7

th

\newline

7

th

\newline

Unit

7

: PC

\newline

Homew

\newline

This is a

2

-page documer

\newline

Directions: If each quadrilateral below is a parallelogram,

\newline

1

.

\newline

\begin{array}{l} M N= \\ K N= \\ m \angle K= \\ m \angle L= \\ m \angle M= \end{array}

\newline

3

. Given

P Q=24, P S=19, P R=42, T Q=10, m \angle P Q R=10

Parallelogram Properties: Since $PQRS$ is a parallelogram, opposite sides are equal. So, $PQ = RS$ and $PS = QR$ . $\newline$ Calculation: $PQ = 24$ , so $RS = 24$ . $PS = 19$ , so $QR = 19$ .
Diagonals Bisect Each Other: The diagonals of a parallelogram bisect each other. So, $PR$ is bisected into $PT$ and $TR$ , and $QS$ is bisected into $QT$ and $TS$ . $\newline$ Calculation: $PR = 42$ , so $PT = TR = \frac{42}{2} = 21$ .
Finding QS: Since $TQ$ is given as $10$ , and $QT$ is half of $QS$ , $QS = 2 \times TQ$ . $\newline$ Calculation: $QS = 2 \times 10 = 20$ .
Calculating Angle Q: Now, we can find the measure of angle Q using the fact that the sum of the angles in a triangle is $180$ degrees. Triangle PQR has angles PQR, QPR, and PRQ. $\newline$ Calculation: $m/\angle PQR = 10$ degrees, $m/\angle QPR = 90$ degrees (since PS is perpendicular to QR), so $m/\angle PRQ = 180 - 10 - 90 = 80$ degrees.
Correction: We made a mistake in the previous step. We assumed that $PS$ is perpendicular to $QR$ without any information given that suggests a right angle. We cannot determine $m/\angle QPR$ is $90$ degrees.

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