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Nadya [2.5K]
3 years ago
15

A painter can paint 3 walls in one hour. Which graph models a relationship with the same unit rate?​

Mathematics
1 answer:
Irina18 [472]3 years ago
5 0

Answer:

the bottom faded looking one

Step-by-step explanation:

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I'm stuck on this one:( help me please ​
ZanzabumX [31]

Answer:

Step-by-step explanation:

f-1(x) is the inverse of f(x)

to find the inverse let’s set the equation up like this

y = 4x -2

now we’ll switch x and y

x = 4y -2

and solve for y

x + 2 = 4y

x/4 + 2/4 = y

the inverse is

y = x/4 + 1/2

this equation is written in y = mx + b form

where b = y intercept

so the y intercept is 1/2

6 0
3 years ago
Help is needed as we speak/write.
san4es73 [151]

Answer:

A= 4 miles

B=  d=2h+4

C= 4 hours

Step-by-step explanation:

4 0
3 years ago
What is the value of log625^5? A: -4 B:-1/4 C:1/4 D:4
shtirl [24]

Answer:

The value of log_{625}(5) is \frac{1}{4}

Step-by-step explanation:

We want to evaluate log_{625}(5).


The base of this logarithm is 625 and the number is 5.


We need to express the number 5 as the base 625 raised to a certain index.


This implies that, log_{625}(5)=log_{625}(625^{\frac{1}{4}})


Recall now that,

log_a(m^n)=nlog_a(m).


We apply this property to obtain,

log_{625}(5)=\frac{1}{4}log_{625}(625)


Recall again that,


log_a(a)=1,a\ne0\:or\:1.This implies that,


log_{625}(5)=\frac{1}{4}(1)


log_{625}(5)=\frac{1}{4}


The correct answer is C.



3 0
3 years ago
Read 2 more answers
Need help solving this .
vodka [1.7K]
D. 37.5% is the correct answer for your question
8 0
4 years ago
Read 2 more answers
A fire station is to be located along a road of length A, A < q. If fires occur at points uniformly chosen on (0, A), where s
Shtirlitz [24]

Answer:

Step-by-step explanation:

Given that X is uniform in the interval (0,A)

X is continuous since it represents the distance

A fire station is to be located along a road of length A, A < q. If fires occur at points uniformly chosen on (0, A), we  should find where the station be located so as to minimize the expected distance from the fire

Distance can be taken as absolute values here as either side is the same.

E(|x-a|] is to be minimum

E(|x-a|)=E(x-a) for 0<x<a

         =E(a-x), for a<x<A

Using integral we find this value

E(|x-a|)=\int\limits^a_0 {x-a} \, dx +=\int\limits^A_a -{x-a} \, dx\\f(a)=\frac{a^2}{2} -\frac{(A-a)^2}{2}

f(a)=\frac{a^2}{2} +\frac{(A-a)^2}{2} \\f'(a) = a-(A-a)\\f"(a) =2

Using calculus we find that f" is positive so when f' =0 we get solution

f'(a) =0 gives

a=\frac{A}{2}

Hence fire station to be located at the mid point of 0 and A

8 0
4 years ago
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