All categories

3 years ago

what is the equation of a line passes through the points(0,5) and (4,8) answer it in slope-intercept form

Mathematics

2 answers:

solmaris [256]3 years ago

6 0

Your answer’s y=3/4x+5 . yw?

alexdok [17]3 years ago

4 0

Answer:

$y=\frac{3}{4} x+5$

Step-by-step explanation:

Substitute into the slope formula

$m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

$m=\frac{8-5 }{4-0 }$

$m=\frac{3}{4}$

Substitute slope and one point (doesn't matter which) into slope intercept form (y=mx+b) to solve for b

Write the equation

I hope this helps :)

You might be interested in

The new balance of a charge account is $145.00. The minimum payment is $20.00. If the interest rate is 1.5%, how much will the n

juin [17]

The balance is $145 but after one makes he minimum payment of $20, there is $125 left to pay. However interest is added before the next month’s bill is generated.

Remember that percent means out of 100. 1.5% means 1.5/100 = .015

To find 1.5% of 125 we multiply (.015)(125) = 1.875 But as this is money the interest is rounded to two decimal places and so is $1.88

We add this to 125 to get next month’s balance which is 125 + 1.88 = $126.88

3 0

4 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

3 0

2 years ago

I need help with math

nataly862011 [7]

Answer:

is that middle school or what

7 0

3 years ago

DO IN 30 MINUTES <br> Pls help :(

Zina [86]

If you can’t answer all of them on time I recommend you to take a picture then write down the work that way you can say to the teacher these are the answers for the ones I left blank

6 0

3 years ago

some orange paint is made by mixing 4 mL red paint and 15 mL of yellow paint. How much red paint should be mixed with 100 mL of

Korolek [52]

The answer is 6. 6666ml

6 0

3 years ago