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lukranit [14]
3 years ago
5

How do you solve 4x+5y=20?​

Mathematics
1 answer:
gtnhenbr [62]3 years ago
8 0

Answer:

are you solving for y or x?

edit:

y= -4/5x + 4

x=-5/4y + 5

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Perpendicular to y=4x-7; contains the point (4,3)
san4es73 [151]
Okay, so a general rule for finding perpendicular lines in the form of y = mx + b is y = (-1/m) + b.
First, let's ignore b (-7) because we're going to find that later.
A perpendicular line to y = 4x + b is y = -1/4x + b.
Alright, so now let's plug in the values. They are in the form of (x,y), so let's plug them in accordingly.

3 = -1/4(4) + b
3 = -1 + b
b = 4
y = -1/4x + 4
So a line perpendicular to y = 4x - 7 is y = -1/4x + 4.
5 0
3 years ago
What is the constant rate of change in this?
valentina_108 [34]

Answer:

function

Step-by-step explanation:

hahahahahahahbasta yan ata ang sagot

8 0
3 years ago
The relative growth rate for a certain type of mutual fund is 15% per year. An account is opened with a balance of $3,000. How m
wlad13 [49]

We have been given that an account is opened with a balance of $3,000 and relative growth rate for a certain type of mutual fund is 15% per year.  

In order to tackle this problem we have to find the value of mutual fund after 5 years. For our purpose we will use compound interest formula.

A=P(1+r)^{t} ,where A= amount after t years, P= principal amount, r= interest rate (decimal) and t= number of years.

After substituting our given values in above formula we will get

A=3000(1+.15)^{5}

Now we will solve for A

A=3000(1+.15)^{5}=3000(1.15)^{5}\\ A=3000\cdot (2.011357187)=6034.07

Therefore, after 5 years mutual fund is worth $6034.07.


8 0
4 years ago
Find the area between the two functions
Inga [223]

The area between the two functions is 0

<h3>How to determine the area?</h3>

The functions are given as:

f₁(x)= 1

f₂(x) = |x - 2|

x ∈ [0, 4]

The area between the functions is

A = ∫[f₂(x) - f₁(x) ] dx

The above integral becomes

A = ∫|x - 2| - 1 dx (0 to 4)

When the above is integrated, we have:

A = [(|x - 2|(x - 2))/2 - x] (0 to 4)

Expand the above integral

A = [(|4 - 2|(4 - 2))/2 - 4] - [(|0 - 2|(0 - 2))/2 - 0]

This gives

A = [2 - 4] - [-2- 0]

Evaluate the expression

A = 0

Hence, the area between the two functions is 0

Read more about areas at:

brainly.com/question/14115342

#SPJ1

5 0
2 years ago
PLS HELP ME WITH 3!<br><br> THANK YIU SO MUCH!!<br><br><br> (Random answers gets moderated.)
kirill115 [55]
6 {x}^{2} - 7x - 5 \\ 6 {x }^{2} + 3x - 10x - 5
3x(2x + 1) - 5(2x + 1) \\ (3x - 5)(2x + 1)
8 0
3 years ago
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