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Advocard [28]
3 years ago
7

A line passes through point (-8,-6) and has a slope of

Mathematics
1 answer:
Viefleur [7K]3 years ago
4 0

Answer:

5x - 2y= -28

Step-by-step explanation:

Since the y= mx + b format would be: y= 5/2x + 14

Standard form is: 5x - 2y= -28

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Instructions: Use the ratio of a 30-60-90 triangle to solve for the variables.
Pepsi [2]

Answer:

x=20

y=10\,\sqrt{3}

Step-by-step explanation:

In the triangle we have the following trigonometric equations:

tan(60^o)=\frac{opposite}{adjacent} \\\sqrt{3} =\frac{y}{10} \\y=10\,\sqrt{3}

and also:

cos(60^o)=\frac{adjacent}{hypotenuse}\\\frac{1}{2} =\frac{10}{x}\\x=20

8 0
3 years ago
Help please I get y=mx+b but I'm not sure how to do this
Fofino [41]

(3,4)  (9,7)

to find the slope

m = (y2-y1)/(x2-x1)

  = (7-4)/(9-3)

3/6

1/2

The slope is 1/2

point slope form

y-y1 = m(x-x1)

y-4 = 1/2(x-3)

distribute

y-4 = 1/2x -3/2

add 4 to each side

y = 1/2 x -3/2 + 4

y = 1/2 x -3/2 + 8/2

y = 1/2 x + 5/2

this is in slope intercept form

8 0
3 years ago
Read 2 more answers
Last one plz help me!
Mashutka [201]

Answer:

a) no real solutions

Step-by-step explanation:

the discriminant (b² - 4ac) is negative which indicates no real solutions

3 0
3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
Does1/3 &amp; 7/21 form a proportion
lesantik [10]
Because 1 * 21 = 3 * 7 ( 21  = 21 ) , 1/3&7/21 form a proportion.
8 0
3 years ago
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