If the two acute angles of a right triangle are congruent, then what is the measure of each acute angle?

1 answer:

7 0

Each one is equal to 90°/2 = 45°

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yanalaym [24]

The equation is most likely y=8t+x

7 0

3 years ago

anygoal [31]

Answer:

Parallel

Step-by-step explanation:

Parallel lines are 2 coplanar lines that dont intersect at any point

4 0

2 years ago

nataly862011 [7]

To evaluate a function at a given input, you have to substitute every occurrence of the variable with that particular value.

So, for the first function, you have

$T(x) = -x-2 \implies T(-4) = -(-4)-2 = 4-2 = 2$

The name of the variable is obviously irrelelvant, so the same goes for the second function:

$w(x) = 14-6x \implies w(-3) = 14-6\cdot(-3) = 14 + 18 = 32$

3 0

2 years ago

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$