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3 years ago

Find the value of x. Round the length to the nearest tenth. The diagram is not drawn to scale.

Mathematics

1 answer:

Gekata [30.6K]3 years ago

4 0

Answer:

x ≈ 2879.4 m

Step-by-step explanation:

Measure of ∠DAC = 10°

Measure of ∠DAB = 90°

Measure of ∠BAC = 90° - 10° = 80°

By applying cosine rule in ΔABC,

cos(80°) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

= $\frac{AB}{AC}$

0.17365 = $\frac{500}{x}$

x = $\frac{500}{0.17365}$

x = 2879.36

x ≈ 2879.4 m

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What is the value of x in the solution to this system of equations?

liberstina [14]

Answer:

x = 11

Step-by-step explanation:

Given the expression;

5x+2y=67 ...1

x=3y−7 ....2

Substitute 2 into 1

5(3y-7)+2y = 67

15y - 35 + 2y = 67

17y = 67+35

17y = 102

y = 102/17

y =6

Recall that x = 3y - 7

x = 3(6) - 7

x = 18 - 7

x = 11

Hence the value of x is 11

6 0

2 years ago

In your biology class, your nal grade is based on several things: a lab score, scores on two major tests, and your score on the

Mariana [72]

Complete Question

In your biology class, your final grade is based on several things: a lab score, scores on two major tests, and your score on the final exam. There are 100 points available for each score. However, the lab score is worth 25% of your total grade, each major test is worth 22:5%, and the final exam is worth 30%. Compute the weighted average for the following scores: 92 on the lab, 81 on the first major test, 93 on the second major test, and 85 on the final exam. (Enter your answer to one decimal place)

Answer:

The weighted average is $W = 87.65$

Step-by-step explanation:

From the question we are told that

The worth of the lab score is 25% of total grade

The worth of each major test is 22.5% of total grade

The worth of final exam is 30% of total grade

The score on lab is 92

The score on first major test 81

The score on second major test 93

The score on final exam is 85

Generally the weighted score is mathematically represented as

$W = 0.25 * 92 + 0.225* 81 + 0.225* 93 +0.30 * 85$

=> $W = 0.25 * 92 + 0.225* 81 + 0.225* 93 +0.30 * 85$

=> $W = 87.65$

3 0

3 years ago

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

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: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$