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2 years ago

What is the equation of the following line written in slope-intercept form?

Mathematics

1 answer:

nasty-shy [4]2 years ago

6 0

SOLUTION:

To begin with, let's establish that the formula of this line is in slope-intercept form as follows:

y = mx

The formula for this line isn't:

y = mx + b

This is as this line doesn't have a y-intercept ( b ) as it passes through the origin instead. This means that ( b ) would be rendered useless in this formula as it would just bring us back to the y = mx formula as displayed below:

y = mx + b
y = mx + 0
y = mx

Moving on, for ( m ), we need to find the gradient of the line as displayed below:

m = gradient
m = rise / run
m = 10 / 2
m = 5

Now, we must simply substitue ( m ) into the formula in order to obtain the equation for this line as displayed below:

y = mx
y = 5x

Therefore, the answer is:

A. y = 5x

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Which answer is the solution set in simpliest form 4x-1/3=7/1

Travka [436]

4x - 1/3 = 7/1
+1/3 +1/3
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4x = 7.3 (repeating)

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x = 7.3 (repeating) / 4

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2 years ago

Describe how to redraw a scale drawing with a new scale

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2 years ago

A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500

Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking, $\frac{d(w)}{dt}$ = 3 ft/sec

Rate at which the man is walking, $\frac{d(m)}{dt}$ = 2 ft/sec

Collective rate of both, $\frac{d(m+w)}{dt}$ = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min = 20 min

Now,

Distance traveled by man and woman are $m$ and $w$ ft respectively.

⇒ $m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft$

⇒ $w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\ ft$

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate $\frac{dh}{dt}$ .

Lets calculate h.

Applying Pythagoras formula.

⇒ $h^2=(m+w)^2+500^2$

⇒ $h=\sqrt{(2400+2700)^2+500^2} = 5124.45$

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ $h^2=(m+w)^2+500^2$

⇒ $2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt} + \frac{d(500)}{dt}$

⇒ $\frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}$ ...as $\frac{d(500)}{dt}= 0$

⇒ Plugging the values.

⇒ $\frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}$ ...as $\frac{d(m+w)}{dt} = 5$ ft/sec

⇒ $\frac{d(h)}{dt} =4.9761$ ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

3 0

2 years ago

Help pls FAST !! Plsss

Lorico [155]

Answer:

$\sqrt{17}$ , 4.23 , 9/2

Step-by-step explanation:

$\sqrt{17} = 4.1231\\$

9/2 = 4.5

4.23

7 0

2 years ago

Given that (ax^2 + bx + 3) (x+d) = x^3 + 6x^2 + 11x + 12, it asks a + 2b - d = ?

astraxan [27]

Answer:

a+2b-d=1, 3, 5, 7

Step-by-step explanation:

(ax^2+bx+3)(x+d)

ax^3+bx^2+3x+adx^2+bdx+3d

ax^3+bx^2+adx^2+3x+bdx+3d=x^3+6x^2+11x+12

ax^3=x^3, a=1

bx^2+adx^2=6x^2

x^2(b+ad)=6x^2

b+ad=6

b+(1)d=6

b+d=6

------------

3x+bdx=11x

x(3+bd)=11x

3+bd=11

-----------------

b=6-d

3+(6-d)d=11

3+6d-d^2=11

3-11+6d-d^2=0

-8+6d-d^2=0

d^2-6d+8=0

factor out,

(d-4)(d-2)=0

zero property,

d-4=0, d-2=0

d=0+4=4,

d=0+2=2

b=6-4=2,

b=6-2=4.

------------------

a+2b-d=1+2(2)-2=1+4-2=5-2=3

-------------------

a+2(4)-4=1+8-4=9-4=5

-----------------------

a+2(2)-4=1+4-4=5-4=1

-----------------------

a+2(4)-2=1+8-2=9-2=7

5 0

2 years ago

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