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3 years ago

What is 3x+5y=40 in slope intercept form

Mathematics

1 answer:

cupoosta [38]3 years ago

3 0

Hey, hope you're doing great today :)

Let's learn about what we need to learn to solve our equation.

We need to make this into a y=mx+b form.

That's the main thing you need to understand while doing this.

Now let's get onto the problem!

This is to much big of an equation to set up, so we need to change some things up.

We need to make one thing into a fraction.

It should look like y = -3/5x + 8.

This is in correct y = mx + b form.

Hopefully that clears things up :)

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Which is the correct word form of 8.16

Aleksandr-060686 [28]

The correct word form of 8.16 is b. eight and sixteen hundredths

Since the number is 8.16, we have 8.16 = 8 + 0.16 = 8 + 16/100

Since we have 8 in the units place, it is pronounced eight.

Also, we have our decimal part 0.16 = 16/100 which is pronounced as sixteen hundredths.

So, combining both we have eight and sixteen hundredths.

So, the correct word form of 8.16 is b. eight and sixteen hundredths

Learn more about decimals here:

brainly.com/question/8829988

7 0

3 years ago

What is y - 4/5x = 1 y-x=1

madam [21]

Go to a math calculator and type this in then press graph and it will give you your answer

7 0

3 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

What is the mean of the following data values? 53,71,89,10,87

enot [183]

Answer:

Step-by-step explanation:

Mean=Sum of all values/number of observation

=53+71+89+10+87/5

=310/5

=62

7 0

3 years ago

Get the algebraic expressions in the following cases using variables constants and arthmetic a subtract 1)Subtract 25 from the s

Troyanec [42]

Answer:

1.) (a + b) - 25

2.) sqrt(y) - sqrt(x)

Step-by-step explanation:

when they use the word 'from,' you know that it will be the second term, also dont forget your parentheses

5 0

3 years ago