A: 10 B: 70 C: 30 D: 90

Write the standard form of the line that passes through the point (-2, 4) and is parallel to x - 2y = 6.

Radda [10]

First calculate the function of x-2y=6:
x-2y=6 | - x
-2y=6-x | ÷ (-2)
y=0.5x-3

Now you have the function. Because the slope you search is parallel to this one, the slope is the same (in this case 0.5). A slope of 0.5 means for 2 steps to the right go one up. To define the function you are searching you have to find out the y intercept. For this take your point (-2|4). Now go as many steps to the right until you reach the y intercept or x=0 in this case you need 2 steps. So for 2 steps right you go one up. Now you have the y intercept (5).
The resulting function is: f(x) = 0.5+6.

6 0

3 years ago

Write the equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x

Solnce55 [7]

Answer:

The equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x is $\mathbf{y=-\frac{2}{3}x+3 }$

Step-by-step explanation:

We need to Write the equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x

The equation in slope-intercept form is: $y=mx+b$ where m is slope and b is y-intercept.

Finding Slope:

The both equations given are parallel. So, they have same slope.

Slope of given equation y= - 2/3x is m = -2/3

This equation is in slope-intercept form, comparing with general equation $y=mx+b$ where m is slope , we get the value of m= -2/3

So, slope of required line is: m = -2/3

Finding y-intercept

Using slope m = -2/3 and point (-3,5) we can find y-intercept

$y=mx+b\\5=-\frac{2}{3}(-3)+b\\5=2+b\\ b=5-2\\b=3$

So, we get b = 3

Now, the equation of required line:

having slope m = -2/3 and y-intercept b =3

$y=mx+b\\y=-\frac{2}{3}x+3$

The equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x is $\mathbf{y=-\frac{2}{3}x+3 }$

5 0

3 years ago

Read 2 more answers

An object is launched at 29.4 meters per

Lerok [7]

Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

The height of the platform from which the object is launched, h = 34.3 meter

The equation for the height of the object as a function of time 'x' is given as follows;

f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

4 0

3 years ago

Lateral and surface area of prisms?

nekit [7.7K]

Answer:

Lateral surface area is the round part/the middle part of the cylinder, not the bases

Step-by-step explanation:

6 0

3 years ago

HELP PRECALC DO NOT UNDERSTAND WILL GIVE BRAINLIEST

Olegator [25]

Answer:

the lower right matrix is the third correct choice

Step-by-step explanation:

Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).

Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.

__

We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.

The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...

$\left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]$

Already, we see that the second row matches that in the lower right matrix.

The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.

The correct choices here are the two you have selected, and the lower right matrix.

3 0

4 years ago