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goldfiish [28.3K]
3 years ago
10

What is the equation of a line with a slope of 4 that passes through point (-2,5)

Mathematics
1 answer:
12345 [234]3 years ago
8 0

Point slope form is:  y - y1 = m(x - x1)

So we plug in the point we were given, and use the slope, m:

 

y -5 = 4(x - 2) ANSWER

 

Now: y - 5 = 4x - 8,

 

so:         y = 4x - 8 +5  

 

so:        y = 4x - 3 ANSWER  

                 (SLOPE INTERCEPT FORM IS: y =mx + b)    

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What additional piece of information must be known in order to calculate the volume of the cylinder below?
Brilliant_brown [7]

Answer:

π

Step-by-step explanation:

the formula is V = πr2h

pie times radius to the second power times hight

6 0
2 years ago
15*PLEASE PLEASE PLEASE ANSWER VERY FAST AND CORRECT TYSM!
lapo4ka [179]

-2/15=13/15

Step-by-step explanation:

2/3-4/5=13/15

2/15=13/15

4 0
3 years ago
Find the lengths of the missing side . Simplify all radicals !!!<br> help mee!!!!!!
larisa86 [58]

Answer:

e = 13\sqrt{2}

f = 13\sqrt{2}

Step-by-step explanation:

The ∆ given is an isosceles ∆ with a right angle measuring 90°, and two congruent angles measuring 45° each.

Using trigonometric ratio formula, we can find the lengths of the missing side as shown below:

Finding e:

sin(\theta) = \frac{opp}{hyp}

sin(\theta) = sin(45) = \frac{\sqrt{2}}{2}

hyp = 26

opp = e = ?

Plug in the values into the formula

\frac{\sqrt{2}}{2} = \frac{e}{26}

Multiply both sides by 26

\frac{\sqrt{2}}{2}*26 = \frac{e}{26}*26

\frac{\sqrt{2}}{2}*26 = e

\frac{\sqrt{2}}{1}*13 = e

13\sqrt{2} = e

e = 13\sqrt{2}

Since side e is of the same length with side f, therefore, the length of side f = 13\sqrt{2}

3 0
3 years ago
What is the area of the regular polygon shown below?
Alexxx [7]
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7 0
2 years ago
Which equation has a graph that is perpendicular to the graph of -x + 6y = -12?
serious [3.7K]

Answer:

c) 6x + y = -52  is required equation perpendicular to the given equation.

Step-by-step explanation:

If the equation is of the form    : y = mx  + C.

Here m = slope of the equation.

Two equations are said to be perpendicular if the product of their respective slopes is -1.

Here, equation 1 :  -x + 6y = -12

or, 6y = -12  + x

or, y = (x/6)  - 2

⇒Slope of line 1 = (1/6)

Now, for equation 2  to be  perpendicular:

Check for each equation:

a. x + 6y = -67       ⇒  6y = -67  - x

or, y = (-x/6)  - (67/6)      ⇒Slope of line 2 = (-1/6)

but \frac{1}{6} \times \frac{-1}{6}  \neq -1

b. x - 6y = -52   ⇒  -6y = -52  - x

or, y = (x/6)  + (52/6)      ⇒Slope of line 2 = (1/6)

but \frac{1}{6} \times \frac{1}{6}  \neq -1

c. 6x + y = -52    

or, y =y = -52  - 6x      ⇒Slope of line 2 = (-6)

\frac{1}{6} \times (-6)  =  -1

Hence, 6x + y = -52  is required equation 2.

d. 6x - y = 52  ⇒  -y = 52  - 6x

or, y = 6x   - 52      ⇒Slope of line 2 = (6)

but \frac{1}{6} \times 6  \neq -1

Hence,  6x + y = -52  is  the  only required equation .

3 0
3 years ago
Read 2 more answers
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