All categories

3 years ago

Work out m and c for the line: y = 4 x − 1

Mathematics

2 answers:

vaieri [72.5K]3 years ago

6 0

Answer:

m=4 and c=-1

Step-by-step explanation:

I'm assuming that m is the slope and c is the y-intercept. (If this is not the case, please tell me and I'll fix my answer).

The equation is already in the form y=mx+c so you just take the numbers from the problem y=4x-1.

leva [86]3 years ago

6 0

Answer:

m = 4 c=-1

Step-by-step explanation:

The slope form is y = mx + b.

M is the slope which is 4.

B is the y=intercept which is -1.

You might be interested in

What is the answer to this problem (x+3)(x+5)

klasskru [66]

Easy, showing work,

using the method FOIL (First,Outer,Inner, Last)

Take, (x+3)(x+5) and multiply,

First,

x*x= $x^{2}$

Outer,

x*5=5x

Inner,

3*x=3x

Last,

3*5=15

Add the terms together,

$x^{2}$ +8x+15 ( $x^{2}$ +5x+3x+15)

Thus, your answer.

4 0

3 years ago

Read 2 more answers

The equation d= -6t + 10 represents the distance d, in miles, Ralph walks from the library to his house in t hours. His sister,

Artist 52 [7]

A: 2.5
Don’t know if I’m right

7 0

4 years ago

From a point A that is 8.20 m above level ground, the angle of elevation of the top of a building s 31 deg 20 min and the angle

Mariulka [41]

Answer:

30.11 meters ( approx )

Step-by-step explanation:

Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

Point A is 8.20 m above level ground,

So, the height of the building = ( x + 8.20 ) meters,

Now, 1 degree = 60 minutes,

⇒ $1\text{ minute } =\frac{1}{60}\text{ degree }$

$20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}$

$50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}$

By the below diagram,

$tan ( 12^{\circ} 50') = \frac{8.20}{y}$

$tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}$

$tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}$

$\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}$

Now, again by the below diagram,

$tan (31^{\circ}20')=\frac{x}{y}$

$tan(31+\frac{1}{3})=\frac{x}{y}$

$\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91$

Hence, the height of the building = x + 8.20 = 30.11 meters (approx)

8 0

3 years ago

Two objects were lifted by a machine. One object had a mass of 2 kilograms and was lifted at a speed of 2m/sec. the other had a

OLga [1]

Sadly, after giving all the necessary data, you forgot to ask the question.
Here are some general considerations that jump out when we play with
that data:

<em>For the first object:</em>
The object's weight is (mass) x (gravity) = 2 x 9.8 = 19.6 newtons
The force needed to lift it at a steady speed is 19.6 newtons.
The potential energy it gains every time it rises 1 meter is 19.6 joules.
If it's rising at 2 meters per second, then it's gaining 39.2 joules of
potential energy per second.
The machine that's lifting it is providing 39.2 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(2)(4) = 4 joules.

<em>For the second object:</em>
The object's weight is (mass) x (gravity) = 4 x 9.8 = 39.2 newtons
The force needed to lift it at a steady speed is 39.2 newtons.
The potential energy it gains every time it rises 1 meter is 39.2 joules.
If it's rising at 3 meters per second, then it's gaining 117.6 joules of
potential energy per second.
The machine that's lifting it is providing 117.6 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(4)(9) = 18 joules.

If you go back and find out what the question is, there's a good chance that
you might find the answer here, or something that can lead you to it.

4 0

3 years ago

(NEED ASAP PLEASE ILL MAKE YOU BRAINLIEST ANSWER (don’t give me the 0.2 units cause this is not the right math problem for that)

inysia [295]

The first one would be (-5,5) and the second one is (5,5) or 1.

8 0

3 years ago