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3 years ago

Could you please help me with 5x−4y=16?

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

3 0

Answer:

Rewrite in slope-intercept form.

y = 5/4x + 4

Using the slope-intercept form, the y-intercept is b=4

Step-by-step explanation:

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Approximate the square root of √31 .Round to two decimal places

ella [17]

Answer: 5.57

Step-by-step explanation:

The square root of 31 is 5.56776436 and if you round that 2 decimals places you get 5.57 bc anything higher than 5 gets rounded up to the next number.

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4 years ago

Find the roots of the following factored quadratic equation.<br> y= (2x-9) (7x-4)

Marina86 [1]

Answer:

x = 4.5, 4/7

Step-by-step explanation:

To find the roots of a factored equation you need to set the y value at 0, that way you can solve for x. Once you solve for x you get two quantities, which can be graphed. These two quantities are where the x intercepts are.

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3 years ago

A video store charges $5 per game rental plus $12 to rent the game system. Write an expression for renting the game system and a

spin [16.1K]

Y=5x+12
Y being the total amount and x being the number of games

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3 years ago

In a game of tug of war, your team changes -2 1/2 feet in position every 10 seconds. What is your change in position after 40 se

Airida [17]

The answer is -10

-2 1/2 x 4 = -10

8 0

3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago