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3 years ago

Determine the length of the missing side of the right triangle? ? inches 15 inches 8 inches

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

3 0

Answer:

the missing angle would be 17 inches. I am not exactly sure but if the leg a and leg b were calculated with those 2 numbers then the other side(known as hypotenuse) should be 17 inches

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Sonbull [250]

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3 years ago

there ar 2.54 centimeters in 1 inch. how many cantimeter are in 1 foot? in 1 yard explain you reasoning

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3 years ago

How to solve 3(×-5)-5=23

USPshnik [31]

Step 1: Simplify both sides of the equation.

3(x−5)−5=23

(3)(x)+(3)(−5)+−5=23(Distribute)

3x+−15+−5=23

(3x)+(−15+−5)=23(Combine Like Terms)

3x+−20=23

3x−20=23

Step 2: Add 20 to both sides.

3x−20+20=23+20

3x=43

Step 3: Divide both sides by 3.

3x /3 = 43 /3

x= 43 /3

4 0

3 years ago

Read 2 more answers

Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P'Q'R' has

Whitepunk [10]

Answer:

Please find attached the required plot accomplished with an online tool

Part A:

1/4

Part B:

P''(-1, 0), Q''(0, -1), and R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent

Step-by-step explanation:

Part A:

Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)

Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)

The dimensions of the sides of the triangle are given by the relation;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment

For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);

By substitution into the length equation, we get;

The length of segment PQ = 4·√2

The length of segment PR = 4·√10

The length of segment RQ = 8

The length of segment P'Q' = √2

The length of segment P'R' = √10

The length of segment R'Q' = 2

Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4

Part B:

Reflection of (x, y) across the y-axis gives;

(x, y) image after reflection across the y-axis = (-x, y)

The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;

P'(1, 0) image after reflection across the y-axis = P''(-1, 0)

Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)

R'(-2, -1) image after reflection across the y-axis = R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.