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2 years ago

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find the equation of the linear function which passes through the points (1,-3) and (4,-2) give your answer in y=mx+b form

1 answer:

Tpy6a [65]2 years ago

5 0

To solve this problem, we first have to find the slope of the two given points. To find the slope, we have to use slope intercept form, which is y₂ - y₁ divided by x₂ - x₁. We substitute each variable for its corresponding value, and we'll get -2 - (-3) divided by 4-1. If we simplify the expression, we'll get 1/3. Next, we plug in the slope and one ordered pair into the equation and we get -3 = 1/3 * 1 + b. We should solve for b and we end up with -3 1/3 = b. Therefore, the equation is y = 1/3x - 3 1/3.

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Step-by-step explanation:

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2 years ago

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3 years ago

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fomenos

Answer:

A and B

Step-by-step explanation:

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2 years ago

Solve x'=5t(sqrt(x)) x(0)=1

LekaFEV [45]

Answer:

$2\sqrt{x}=\frac{5t^2}{2}+2$

Step-by-step explanation:

Given: $\frac{\mathrm{d} x}{\mathrm{d} t}=5t\sqrt{x}\,,\, x(0)=1$

Solution:

A differential equation is said to be separable if it can be written separately as functions of two variables.

Given equation is separable.

We can write this equation as follows:

$\frac{dx}{\sqrt{x}}=5t\,dt$

On integrating both sides, we get

$\int \frac{dx}{\sqrt{x}}=\int 5t\,dt$

Formulae Used:

$\int \frac{1}{\sqrt{x}}=2\sqrt{x}\,\,,\,\,\int t\,dt=\frac{t^2}{2}$

So, we get solution as $2\sqrt{x}=\frac{5t^2}{2}+C$

Applying condition: x(0) = 1, we get $C=2$

Therefore, $2\sqrt{x}=\frac{5t^2}{2}+2$

8 0

2 years ago

Quadrilateral BCDE is a rhombus. What is DE?<br> D<br> с<br> 9z<br> E<br> 2+72<br> B

aleksklad [387]

Answer:

DE = 81

Step-by-step explanation:

The sides of a rhombus are congruent , thus

9z = z + 72 ( subtract z from both sides )

8z = 72 ( divide both sides by 8 )

z = 9

Then

BC = 9z = 9 × 9 = 81

Since the sides are congruent, then DE = 81

6 0

2 years ago