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andrey2020 [161]
2 years ago
5

find the equation of a straight line passing through the point (3,3) which is perpendicular to the line y=-1/2x-4

Mathematics
1 answer:
Mars2501 [29]2 years ago
4 0

Answer:

y = 2x - 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - \frac{1}{2} x - 4 ← is in slope- intercept form

with slope m = - \frac{1}{2}

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{-\frac{1}{2} } = 2, thus

y = 2x + c ← is the partial equation

To find c substitute (3, 3) into the partial equation

3 = 6 + c ⇒ c = 3 - 6 = - 3

y = 2x - 3 ← equation of perpendicular line

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4 times x with an exponent of 2 plus 2 times y with an exponent of 4. X=-3 y=2
Digiron [165]

Answer:

Step-by-step explanation:

Your equation is: 4*x^{2} +2 * y^{4}

You also know that x = -3 and y = 2. You can rewrite the equation by substituting x and y with their actual values:

4*(-3)^{2} +2 * 2^{4}

First, you want to do the exponentiation (meaning you want to calculate (-3)^{2} and 2^{4} parts). (-3)^{2} equals 9 and 2^{4} equals 16.

So the equation now is:

4*9+2*16

Now you want to do multiplication. 4*9 equals 36 and 2*16 equals 32. So you're left with:

36+32

Which equals 68.

8 0
2 years ago
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100°

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3 0
2 years ago
Suppose y varies inversely with x. Write an equation if y=-3 when x=17
vodka [1.7K]

Answer:

y = -51 ÷ x

Step-by-step explanation:

The standard format of a inversely proportional equation goes by the format of y = k ÷ x. The question asks us to write an equation if y = -3 when x = 17 and y varies inversely with x.

So we substitute the values y = -3 and x = 17 in the equation y = k ÷ x

K = constant, we have to work this out

-3 = k ÷ 17

-3 × 17 = k

k = -3 × 17

k = -51

k = -51 So then we put this back into the original y = k ÷ x to find our final equation which is y = -51 ÷ x

6 0
3 years ago
A scientist is growing bacteria in a lab for study. One particular type of bacteria grows at a rate of y=2t^2+3t+500. A differen
Leya [2.2K]

ANSWER

Approximately after 15 minutes.

EXPLANATION

The growth rate of the first bacteria is

y = 2 {t}^{2}   + 3t + 500

The growth rate of the first bacteria :

y = 3 {t}^{2}  + t + 300

To find the time that, there will be an equal number of bacteria, we equate the two equation;

3 {t}^{2}  + t + 300 = 2 {t}^{2}   + 3t + 500

3 {t}^{2}  - 2 {t}^{2}   + t  - 3t +  300 - 500 =00

{t}^{2}  - 2t  - 200= 0

We solve for t to get,

t = 15.177

Or

t =  - 13.177

We discard the negative value.

This implies that,

t \approx15

8 0
3 years ago
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