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3 years ago

Find the length of side AB in the right triangle shown.

Mathematics

1 answer:

Bogdan [553]3 years ago

5 0

Side AB is 72 cm long.

Explanation:

To find the length of side AB you have to use this formula:

${a}^{2} + {b}^{2} = {c}^{2}$

*Remember there are two legs (a and b) and a hypotenuse (c) on a triangle. The longest side will be the hypotenuse and the other two will be the legs. For this triangle the hypotenuse is side BC while the legs are sides CA and side AB.

Plug in the numbers into the equation.

${65}^{2} + {b}^{2} = {97}^{2}$
For this triangle, you were given the hypotenuse (c) and one leg (a). To find the last length solve the equation and balance it so only b is left, like this:

$4225 + {b}^{2} = 9409$

$9409 - 4225 = 5184$

Soo...

${b}^{2} = 5184$

Then to get the side length, square root 5184.

$\sqrt{5184} = 72$

So the length of the last side is 72 cm.

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Solve for congruent triangles

Elanso [62]

Answer:

11. B.

12. C.

13. A.

14. D.

Step-by-step explanation:

for 11: we know that angles D and J are congruent from the tick marks, we also know that ∠FKD and ∠LKJ are congruent (vertical angles are congruent) therefore we need the sides between them

for 12: we know that ∠STU and ∠TUG are congruent, we also know that line TU is congruent to TU (reflexive property), therefore we need the angles adjacent to the first angles listed.

for 13: we know that ∠PQR and ∠CQR are congruent, we also know that lines RQ and RQ are congruent (reflexive property), therefore we need the other angles to which line RQ is between.

for 14: we know ∠B is congruent to ∠T and line AB is congruent to line ZY. therefore the angle cannot be connected to lines AB and ZY.

8 0

2 years ago

A company offers four plans for customers who want to rent DVDs and video games. The cost of the plans can be modeled by a linea

Luba_88 [7]

To solve this, we are going to use the slope formula, and the point slope formula.
Slope formula: $m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
where
$m$ is the slope.
$(x_{1},y_{1})$ are the coordinates of the first point.
$(x_{2},y_{2})$ are the coordinates of the first point.
Point-slope formula: $y-y_{1}=m(x-x_{1})$

Plan A. Coordinates of the first point: (1,14) (number of discs, cost). Coordinates of the second point: (2,17). Lets replace those values in our slope formula:
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{17-14}{2-1}$
$m=3$
Now that we have the slope, we can use the point slope formula:
$y-y_{1}=m(x-x_{1})$
$y-14=3(x-1)$
$y-14=3x-3$
$y=3x+11$
Remember that in a linear equation of the form $y=mx+b$ , $m$ is the slope and $b$ is the y-intercept. In our model the y-intercept is the membership fee.
Since $b=11$ in our linear equation, we can conclude that the membership fee is $11

Plan B. Coordinates of the first point (1,12). Coordinates of the second point (2,16).
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{16-12}{2-1}$
$m=4$
$y-y_{1}=m(x-x_{1})$
$y-12=4(x-1)$
$y-12=4x-4$
$y=4x+8$
Membership fee: $8

Plan C. Coordinates of the first point: (1,10). Coordinates of the second point: (2,15).
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{15-10}{2-1}$
$m=5$
$y-y_{1}=m(x-x_{1})$
$y-10=5(x-1)$
$y-10=5x-5$
$y=5x+5$
Membership fee: $5

Part D. Coordinates of the first point: (1,12). Coordinates of the second point (2,21).
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{21-12}{2-1}$
$m=9$
$y-y_{1}=m(x-x_{1})$
$y-12=9(x-1)$
$y-12=9x-9$
$y=9x+3$
Membership fee: $3

We can conclude that the plan that has the smallest one-time membership is <span>Plan D.</span>

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3 years ago

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Alex787 [66]

Answer:

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