Need help please ASAP 50points!!! Show work

I NEED HELP IN SPANISH PLEASE

valentina_108 [34]

I can help you in spanish

7 0

3 years ago

Which equations represent the line that is perpendicular to the line 5x – 2y = -6 and passes through the point

natta225 [31]

Answer: 2x + 5y = - 10, Cy + 4 = (x-5)

Dy - 4 = (x+ 5)

Step-by-step explanation:

Equation of the line

5x - 2y = -6

Conditions for perpendicularity

m1 x m2 = -1

To get m1, rearrange the equation

2y = 5x + 6

y = 5x/2 + 3

n1 = 5/2 and m2 = -2/5

To get C

y = mx +c

-4 = -2 x 5/5 + C

-4 = -2 + C

C = -4 + 2

C = -2

To get the equation of the second line

y = -2x/5 - 2

Multiply through by 5

5y = -2x - 10

2x + 5y = 10.

4 0

3 years ago

Read 2 more answers

There are 14 math teachers and 26 science teachers at school today. What is the ratio of math teachers to math and science teach

marin [14]

Answer:

SEE EXPLANATION

Step-by-step explanation:

Add all amount of teachers together

14+26 = 40

14:40

Simplified: 7/20 or 7:20

Ratios are just like fractions and can usually be put in this format

6 0

2 years ago

Read 2 more answers

Can 444 be divisible by 9

Sati [7]

No not evenly, but it will be 49.3 repeating .

8 0

3 years ago

Read 2 more answers

5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l

sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

7 0

3 years ago