Ask question

Ask question

All categories

2 years ago

5

If 15% of 4x is 20 less than x, what is the value of x?

1 answer:

Bess [88]2 years ago

4 0

The sentence "15% of 4x is 20 less than x"

translates to 0.15 • (4x) = x – 20

Solving that…

0.15(4x) = x - 20 multiplipy 0.15•4

0.6x = x - 20 subtract x from both sides

-0.4x = - 20 divide both sides by -0.4

x = 50

You might be interested in

Tres amigos. se van de viaje y quieren. alquilar un coche durante un fin de semana.

djyliett [7]

Answer:

operaciones : 600×3= 1800÷5=360.

Step-by-step explanation:

y entonces cada amigo tendra que pagar 360 ya que eran 3 amigos y iban a pagar 600 cada uno pero ahora son 5 asi que deben de pagar 360

5 0

2 years ago

WILL GIVR YOU BRAINLEST PLEASE HELP ASAP!

netineya [11]

Sorry the question you typed is kinda messy. But I see you have typed “23* 26” I’m gonna assume you’re asking which postulate or theorem is angle 2 and angle 3, and angle 2 and angle 6. So angle 2 and 3 are vertical angles. Angle 2 and 6 are corresponding angles. If that wasn’t what you wanted, please let me know in the comments so I can understand your problem clearly and solve it.

7 0

2 years ago

12=5x-13x-44. answer it in multi-step equations

fomenos

12=5x-13x-44

combine like terms

12 = -8x -44

add 44 to each side

12+44 = -8x

56= -8x

divide by -8 on each side

56/-8 = x

-7 =x

5 0

2 years ago

Read 2 more answers

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

A. 1 and 8 <br> B. 4 and 6<br> C. 5 and 6<br> D. 1 and 7 <br> Please tell me why

andre [41]

1 and 7 because they have the same angles

6 0

2 years ago