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2 years ago

What is the equation of the line that passes through the point (4,-7) and has a slope of -1/2

Mathematics

1 answer:

-BARSIC- [3]2 years ago

8 0

y = (-1/2)x -5

Because the slope is -1/2

Then, plug in 4 as x, (-1/2)4 = -2

To make it -7, just subtract 5.

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Which statement correctly describes the slope of the linear function that is represented by the data in the table?

777dan777 [17]

Slope=change in y/change in x

one way is t pick 2 points, (x1,y1) and (x2,y2)
the slope is (y2-y1)/(x2-x1)

pick some points
(8,4) and (8,0)
slope=(0-4)/(8-8)=-4/0=undefined
the slope is undefined
so you could say there is no solpe

answer is D

5 0

3 years ago

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Jessica needs to make a necklace for 3 of her friends, including her. she wants to use 4.5 inches of string for each necklace. s

hoa [83]

Answer:

She can make two necklaces with 10 inches. 8 more inches

Step-by-step explanation:

5 0

3 years ago

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wat iz 9999999999999999999999999999999999999999999999999999999+99999999999999999999999999999999999999999999999999999999999999999

choli [55]

Ahbfiubauehfiouabcbaiu

5 0

3 years ago

Given 6(x) = (x+41, what is b(-10)?

prisoha [69]

The value of $b(-10)$ is 31

Explanation:

The given expression is $b(x)=x+41$

We need to determine the value of $b(-10)$

The value of $b(-10)$ can be determined by substituting $x=-10$ in the expression $b(x)=x+41$

Thus, we have,

$b(-10)=-10+41$

Adding the terms, we have,

$b(-10)=31$

Thus, the value is 31.

Therefore, the simplified value of the expression by substituting $x=-10$ in the expression $b(x)=x+41$ is 31.

7 0

3 years ago

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0

1 year ago