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2 years ago

7

Isosceles Right Triangle Reflection to prove ASA Congruence

1 answer:

adell [148]2 years ago

7 0

Answer:

Yes they went through rigid motion

Step-by-step explanation:

The original triangle is the same as the one on the left but the only thing that changed was its place(translation) making it a RIGID MOTION

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What is the common ratio of 2, 10/3, 50/9

mojhsa [17]

The common ratio of 2, 10/3, 50/9 is $\frac{10}{6}$

<u>Solution:</u>

Given, series of elements are $2, \frac{10}{3}, \frac{50}{9}$

We have to find the common ratio of the above given series.

We know that, common ratio of an G.P is division of any number in that series with the previous number of the series.

So, now take $\frac{10}{3}$ and 2

$\text { Then, common ratio }=\frac{\frac{10}{3}}{2}$

$\begin{array}{l}{\text { Common ratio }=\frac{10}{3} \times \frac{1}{2}} \\\\ {\text { Common ratio }=\frac{10}{3 \times 2}=\frac{10}{6}}\end{array}$

Hence, the common ratio of the given series is $\frac{10}{6}$

6 0

2 years ago

Which ordered pairs make the equation true?

Nata [24]

(-4,-2)
You just plug in the numbers from the ordered pair, into your equation. Remember, the formula for ordered pairs are (x,y). :)

6 0

2 years ago

Read 2 more answers

Which one is correct

Alexandra [31]

Answer:

i believe its false good Sir

6 0

2 years ago

Read 2 more answers

Use the quadratic formula to find the solutions to the quadratic equation below 2x2 - 3x + 6 = 0

Katarina [22]

Answer:

$x=\frac{3}{4} +i\frac{\sqrt{39}}{4}, x=\frac{3}{4} -i\frac{\sqrt{39}}{4}\\$

OR

x=\frac{3}{4} ±i\frac{\sqrt{39}}{4}\\

Step-by-step explanation:

6 0

2 years ago

LMN is a right-angle triangle. Angle NLM=90. PQ is parallel to LM. The area of triangle PNQ is 8cm^2. The area of triangle LPQ i

likoan [24]

Answer:

The area of LQM is $48cm^2$

Step-by-step explanation:

Given

Area of PNQ = 8

Area of LPQ = 16

See attachment for triangles

The area of PNQ is calculated as:

$Area = \frac{1}{2} * PQ * PN$

Substitute 8 for Area

$8 = \frac{1}{2} * PQ * PN$

$PQ * PN = 16$

The area of LPQ is calculated as:

$Area = \frac{1}{2} * PQ * LP$

Substitute 16 for Area

$16= \frac{1}{2} * PQ * LP$

From the attachment:

$PN + LP =LN$

Make LP the subject

$LP = LN -PN$

So:

$16= \frac{1}{2} * PQ * (LN -PN)$

We have:

$16= \frac{1}{2} * PQ * (LN -PN)$ and $PQ * PN = 16$

Equate both expressions:

$\frac{1}{2} * PQ *(LN - PN) = PQ * PN$

Divide both sides by PQ

$\frac{1}{2} (LN - PN) = PN$

Multiply both sides by 2

$LN - PN = 2PN$

$LN= 3PN$

Since PNQ is similar to LNM, the following equivalent ratios exist:

$\frac{LM}{PQ} = \frac{LN}{PN}$

Substitute $LN= 3PN$

$\frac{LM}{PQ} = \frac{3PN}{PN}$

$\frac{LM}{PQ} = 3$

$LM = 3PQ$

Area of LQM is:

$Area = \frac{1}{2} * LM * LP$

This gives:

$Area = \frac{1}{2} * 3PQ * LP$

$Area = 3 *\frac{1}{2} *PQ * LP$

Recall that:

$16= \frac{1}{2} * PQ * LP$

So:

$Area = 3 *16$

$Area = 48cm^2$

5 0

2 years ago