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3 years ago

11

What is the equation of the line that is perpendicular to the given line and passes through the point (5,3)?

2 answers:

klasskru [66]3 years ago

7 0

Answer:

4x-5y=5

Step-by-step explanation:

What is the equation of the line that is perpendicular to the given line and passes through the point (5,3)?

4x-5y=5

5x+4y=37

4x+5y=5

5x-4y=8

from the graph given , the slope of a graph becomes

S=y2-y1/(x2-x1)

Slope=10-0/(-8-0)

slope=-10/8

for a line that is perpendicular

perpendicular lines=-1/m

negative inverse of the slope

8/10

equation of the line that passes through the point (5,3) becomes

y-y1=m(x-x1)

y-3=8/10(x-5)

5(y-3)=4(x-5)

5y-15=4x-20

5y=4x-20+15

4x-5y=5 A

MrRissso [65]3 years ago

3 0

4x-5y=5 is the answer i got

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Answer:

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3 years ago

The value of -4*2 is______

sashaice [31]

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3 years ago

Read 2 more answers

0/1 For positive integer n, n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?. What is the value of 4# · 3# · 2# · 1#?

MakcuM [25]

Answer:

331776

Step-by-step explanation:

Since n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?

Then 4# = 4? · (4 − 1)? · (4 − 2)?· 1?

= 4? · 3? · 2?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 4? = 4! · (4 − 1)! · (4 − 2)! · 1! = 4! · 3! · 2! · 1! = 288

Thus, 3? = 3! · (3 − 1)! · 1! = 3! · 2! · 1! = 12

Also, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

and 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 4# = 4? · 3? · 2?· 1? = 288 × 12 × 2 × 1 = 6912

We now find 3#

3# = 3? · (3 − 1)? · 1? = 3? · 2?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 3? = 3! · (3 − 1)! · 1! = 3! · 2! · 1! = 12

Thus, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

and, 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 3# = 3? · 2?· 1? = 12 × 2 × 1 = 24

We now find 2#

2# = 2? · (2 − 1)? · 1? = 2? · 1?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 2# = 2?· 1? = 2 × 1 = 2

We now find 1#

1# = 1? · 1? = 1? · 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

And, 1# = 1? · 1? = 1 × 1 = 1

So, 4# · 3# · 2# · 1#? = 6912 · 24 · 2 · 1? = 331776

7 0

3 years ago

I need help with this question

nata0808 [166]

I don't know if you've learnt this, but there's something called the KFC rule (you use it when dividing fractions), so k basically stands for keep (so you keep the first fraction), f stands for flip (so you flip the second fraction) and c stand for change (you change the division sign to a multiplication sign)

So you get

5/20*11/5

55/100

which is 11/20

so, an answer of 11/20 would make Quintin correct

5 0

3 years ago

Read 2 more answers

Select the two values of x that are roots of this equation x^2-5x+2=0

Nataly [62]

The roots of the equation is x = 4.56 OR x = 0.44

<h3>Quadratic equation</h3>

From the question, we are to determine the roots of the given equation

The given equation is

x² -5x +2 = 0

Using the formula method,

$x =\frac{-b \pm \sqrt{b^{2}-4ac } }{2a}$

In the given equation,

a = 1, b = -5, c = 2

Putting the values into the formula,

$x =\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(2) } }{2(1)}$

$x =\frac{5 \pm \sqrt{25-8} }{2}$

$x =\frac{5 \pm \sqrt{17} }{2}$

$x =\frac{5 + \sqrt{17} }{2}$ OR $x =\frac{5 - \sqrt{17} }{2}$

$x =\frac{5 + 4.12}{2}$ OR $x =\frac{5 - 4.12}{2}$

$x =\frac{9.12}{2}$ OR $x =\frac{0.88}{2}$

x = 4.56 OR x = 0.44

Hence, the roots of the equation is x = 4.56 OR x = 0.44

Learn more on Quadratic equation here: brainly.com/question/8649555

#SPJ1

3 0

2 years ago