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Yanka [14]
3 years ago
7

<2 and <6 form what type of angle pair? A.linear pair B.vertical angles C.complementary angles D.supplementary angles

Mathematics
2 answers:
miss Akunina [59]3 years ago
7 0

Answer:

<2 and <6 are complementary angles

Step-by-step explanation:

<2 and <3 are complementary angles since they add to 90

<3 and <6 are vertical angles so they are equal

<3 = <6

Replace <3 with <6

<2 and <6 are complementary angles

WARRIOR [948]3 years ago
4 0

Answer:

the answer should be C please give brainliest

Step-by-step explanation:

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Ahat [919]
We want to find the equation of the normal line of y=\dfrac{20-x}{3x} at the point P=(x_0,y_0), where y_0=3. First calculate x_0. We have:

y_0=f(x_0)=\dfrac{20-x_0}{3x_0}\\\\\\3=\dfrac{20-x_0}{3x_0}\quad|\cdot3x_0\\\\\\3\cdot3x_0=20-x_0\\\\9x_0=20-x_0\\\\9x_0+x_0=20\\\\10x_0=20\quad|:2\\\\\boxed{x_0=2}

Now, when we know that P=(x_0,y_0)=(2,3) we can write an equation of the normal line as:

\boxed{y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)}

Calculate f'(x_0):

f'(x)=\left(\dfrac{20-x}{3x}\right)'=\left(\dfrac{20}{3x}\right)'-\left(\dfrac{x}{3x}\right)'=\dfrac{20}{3}\cdot\left(\dfrac{1}{x}\right)'-\left(\dfrac{1}{3}\right)'=\\\\\\=\dfrac{20}{3}\cdot\left(-\dfrac{1}{x^2}\right)-0=\boxed{-\dfrac{20}{3x^2}}\\\\\\\\f'(x_0)=f'(2)=-\dfrac{20}{3\cdot2^2}=-\dfrac{20}{3\cdot4}=-\dfrac{20}{12}=\boxed{-\frac{5}{3}}

and the equation of the normal line:

y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\y-3=-\dfrac{1}{-\frac{5}{3}}(x-2)\\\\\\y-3=\dfrac{3}{5}(x-2)\\\\\\y-3=\dfrac{3}{5}x-\dfrac{6}{5}\\\\\\y=\dfrac{3}{5}x-\dfrac{6}{5}+3\\\\\\\boxed{y=\dfrac{3}{5}x+\dfrac{9}{5}}
4 0
3 years ago
Does anyone know this?
kumpel [21]

the answer is the first one when you you graph it on a graph you can see it's perpendicular

8 0
3 years ago
An art student is searching for a rectangular canvas to paint on. His professor requires that both the height and width of the c
Pavlova-9 [17]
H>12 and w>12 however p≤60

p=2(h+w) but give what we have above for h and w

p>48 so p must satisfy the solution set:

48<p≤60  and since p=2(h+w)

48<2(h+w)≤60

24<h+w≤30

So there are infinitely many solutions if h and w are not restricted to integer values...

(h,w) vary from (12,18) to (18,12)  Note that neither endpoints exist, 12 because it is explicitly excluded and 18 because that would make the other dimension 12 which is excluded...

Now if you are just talking integer values, there are only:

(13,17),(14,16),(15,15),(16,14),(17,13)
4 0
3 years ago
1) −6k + 7k 2) 12r − 8 − 12 3) n − 10 + 9n − 3 4) −4x − 10x
crimeas [40]

Step-by-step explanation:

I assume we are simplifying

1) -6k+7k = k

2) 12r-8-12 = 12r-20

3) n-10+9n-3 = 10n-13

4) -4x-10x = -14x

7 0
3 years ago
Trudy is starting a comic book collection. Let Cn represent the number of comic books that Trudy has after n weeks. The recursiv
vitfil [10]

Answer:

(A)

C_n=3+2n

(B)

C_1_0=23

Step-by-step explanation:

(A)

we are given

C_n_+_1=C_n+2

C_1=5

Firstly, we will find few terms

C_2=C_1+2

C_2=5+2

C_2=7

C_3=C_2+2

C_3=7+2

C_3=9

C_4=C_3+2

C_4=9+2

C_4=11

so, we will get terms as

5, 7 , 9 , 11

we can see that this is arithematic sequence

First term =5

common difference =d=7-5=2

now, we can use nth term formula

C_n=C_1+(n-1)d

now, we can plug values

C_n=5+2(n-1)

C_n=5+2n-2

C_n=3+2n

(B)

we can plug n=10

C_1_0=3+2\times 10

C_1_0=23

8 0
3 years ago
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