All categories

3 years ago

Write the equation of a line in slope-intercept form that has a slope of 3/4 and passes through the point (-4, 1).

Mathematics

2 answers:

Zolol [24]3 years ago

6 0

Answer:

A) y equals 0.75 x plus 4

Step-by-step explanation:

Another way to solve this to use the point-slope form:

y - y₁ = m(x - x₁)

Substitute the values of m, x₁, y₁ and convert this into slope-intercept form:

y - 1 = 3/4(x - (-4))
y - 1 = 3/4x + 3
y = 3/4x + 4
y = 0.75x + 4

Correct choice is A

oksian1 [2.3K]3 years ago

3 0

Answer:

y = 3/4 x + 4

y equals 0.75 x plus 4

Step-by-step explanation:

y = 3/4 x + b

1 = 3/4 (-4) + b

1 = -3 + b

b = 4

You might be interested in

What is the distance between (-6, 2) and (8, 10) on a coordinate grid?

Vinil7 [7]

Answer:

2√65

Step-by-step explanation:

The difference in the x-coordinates of the two points is 8-(-6) = 14.

The difference in the y-coordinates of the two points is 10-2 = 8.

These distances can be considered to be the lengths of the legs of a right triangle whose hypotenuse is the line segment between the given points. Then the distance between the points is given by the Pythagorean Theorem as ...

... d = √(14² +8²) = √260 = 2√65 ≈ 16.1245

4 0

3 years ago

Read 2 more answers

What are the missing numbers in the sequence below?

stealth61 [152]

Answer:

8.4 , 7.7

Step-by-step explanation:

- 0.7

7 0

3 years ago

This algebra btw , please help

lana [24]

Answer:

Where is the picture, we need the picture.

Step-by-step explanation:

4 0

3 years ago

Anyone help me. ....

weeeeeb [17]

Answer:

see explanation

Step-by-step explanation:

Using the rules of exponents

All the exponents inside the parenthesis are multiplied by the exponent outside the parenthesis.

Then

$\frac{m^{-7}n^{14} }{m^{-25}n^{40} }$ ÷ $\frac{m^{-21}n^{42} }{m^{-25}n^{40} }$

[ Using the rule $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ ]

= $m^{(-7-14)}$ $n^{(14-(-28)}$ ÷ $m^{(-15-10)}$ $n^{(25-(-15)}$

= $m^{-21}$ $n^{14+28}$ ÷ $m^{-25}$ $n^{25+15}$

= $\frac{m^{-21}n^{42} }{m^{-25}n^{40} }$

= $m^{(-21-(-25))}$ $n^{(42-40)}$

= $m^{(-21+25)}$ n²

= $m^{4}$ n² = $m^{x}$ $n^{y}$ , so x = 4 and y = 2

Then

$m^{x-2y}$ = $m^{4-2(2)}$ = $m^{4-4}$ = $m^{0}$ = 1 ← as required

6 0

3 years ago

Find cos 0 for 0 = 30° The 0 has a weird line through it

Nastasia [14]

Answer: $\frac{\sqrt{3}}{2}$

======================================================

Explanation:

The symbol $\theta$ is the greek letter theta. It's often used in trigonometry for angles.

In this case, $\theta = 30^{\circ}$

So,

$\cos(\theta) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}$

You'll need to use a reference sheet or the unit circle to determine the value of cos(30). Or you could use a 30-60-90 triangle template.

When using the unit circle, look in the upper right quadrant (which is quadrant 1). The angle 30 degrees is in this quadrant. Locate the terminal point and note how $\frac{\sqrt{3}}{2}$ is the x coordinate of the terminal point. This is due to $x = \cos(\theta)$

Side note: $\frac{\sqrt{3}}{2} \approx 0.866025$

6 0

3 years ago