All categories

4 years ago

4(f−6) plz help thank u

Mathematics

2 answers:

german4 years ago

8 0

You will times the F then what ever the F equals, subtract it by 6

Arada [10]4 years ago

3 0

Multiply the y by 4 then the -6 by 4

4f - 24

You might be interested in

To motorists leave from the same point at the same time, one traveling east at 36 MPH and the other traveling west at 44 MPH. Ho

Ber [7]

This is a very good question to know about. People who make tests really like it.

The way to solve it is to make the slower car stop and use the difference in speed = to the rate. then do it as an ordinary d = r*t problem.

The difference in speeds is 44 - 36 = 8 miles/ hour.
d = 260 miles
r = 8 mph
t = ???

t = d/r
t = 260/8
t = 32.5 hours.

Thats 32 1/2 hours.

3 0

4 years ago

What is the least natural number which when divided by 3,5,6,8,10 and 12 leaves in each case remainder 2 but when divided by 13

bazaltina [42]

Answer:

962

Step-by-step explanation:

MMM for step by step its kinda just plug and check

962/3 = 320 R2

962/6 = 160 R2

962/8 = 120 R2

962/10 = 96 R2

962/12 = 80 R2

and finally 962/13 = 74

4 0

2 years ago

During a bike challenge riders have to collect serious colored ribbons each 1/2 mile they collect a red ribbon each three 1/8 mi

hjlf

You need to give a little more detail. I don't understand what you are asking.

3 0

4 years ago

If V is the midpoint of and W is the midpoint of , then what is VS?

lilavasa [31]

Answer:

its D, i just took the test lol

Step-by-step explanation:

3 0

4 years ago

Evaluate integral _C x ds, where C is

borishaifa [10]

Answer:

a. $\mathbf{36 \sqrt{5}}$

b. $\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

Step-by-step explanation:

Evaluate integral _C x ds where C is

a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)

i . e

$\int \limits _c \ x \ ds$

where;

x = t , y = t/2

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt}= \dfrac{1}{2}$

and t varies from 0 to 12.

we all know that:

$ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \ \ dt$

∴

$\int \limits _c \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt$

$= \int \limits ^{12}_{0} \ \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2}) \ dt$

$= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0$

$= \dfrac{\sqrt{5}}{4}\times 144$

= $\mathbf{36 \sqrt{5}}$

b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)

Given that:

x = t ; y = 3t²

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt} = 6t$

$ds = \sqrt{1+36 \ t^2} \ dt$

Hence; the integral _C x ds is:

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

Let consider u to be equal to 1 + 36t²

1 + 36t² = u

Then, the differential of t with respect to u is :

76 tdt = du

$tdt = \dfrac{du}{76}$

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145

Thus;

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

$\mathtt{= \int \limits ^{145}_{0} \sqrt{u} \ \dfrac{1}{72} \ du}$

$= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}$

$\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}$

$\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

5 0

4 years ago