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svetlana [45]
3 years ago
11

PLEASE HELP THIS IS A LINEAR EQUATIONS

Mathematics
2 answers:
Stells [14]3 years ago
8 0
D or B I’m not sure
aalyn [17]3 years ago
7 0
Linear equations can be written different ways such the form y=mx+b, where m is the slope of the line and b is the y intercept.
We can substitute our given coordinates into the equations to find out which one is correct.
A:
y =  \frac{2}{3} x + 4
y =  (\frac{2}{3}  \times  - 3) + 4 \\   y= \frac{ - 6}{3}  + 4 \\   y= - 2 + 4 \\  y= 2
For this equation, when the value of x is - 3, y is 2. This means that A is the correct equation as it passes through the given coordinate (-3,2), and has a slope (m) of 2/3.
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Solve the equation. Check your solution.<br> 6x + 4x - 9 = 41<br> x=L
Aleksandr [31]
In this equation x would equal 5
5 0
3 years ago
1. The hypotenuse of a right triangle measures 2 sqrt 15 centimeters and its shorter leg measures 2 sqrt 6
Dominik [7]

Answer:

<h2>50.8°</h2>

Step-by-step explanation:

First, we need to use the Pythagorean Theorem to find the third side. We will then use the law of cosines to find the angle.

2√15² = 2√6² + b²

60 = 24 + b²

Subtract 24

36 = b²

b = 6

Now, that we have all three sides, we will use the law of cosines to figure out the angle.

c² = a² + b² - 2abCosC°

Substitute the values in.

6² = 2√15² + 2√6² - 2(2√15)(2√6)CosC°

36 = 60 + 24 - 2(37.947331922)CosC°

36 = 60 + 24 - 75.894663844CosC°

36 = 84 - 75.894663844CosC°

Subtract 84

-48 = -75.9CosC°

0.63241106719 = CosC°

Arccos (0.63241106719) = 50.77176844°

The largest acute angle is approximately 50.8°

8 0
3 years ago
Read 2 more answers
What is x minus 3 times x plus 2
Marysya12 [62]

Answer:

Your expression would be -2x+2.

5 0
3 years ago
Read 2 more answers
Ricardo has a marble collection with 54 blue marbles,24 red marbles, and 18 yellow marbles into equal rows/how many marbles can
vazorg [7]
Its 2 because you have to find the GCF


7 0
3 years ago
A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe
qaws [65]

The projectile's horizontal and vertical positions at time t are given by

x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t

y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2

where g=9.8\dfrac{\rm m}{\mathrm s^2}. Solve y=0 for the time t it takes for the projectile to reach the ground:

30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s

In this time, the projectile will have traveled horizontally a distance of

x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m

The projectile's horizontal and vertical velocities are given by

v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ

v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about \sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}.

7 0
3 years ago
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