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3 years ago

What angle is supplementary to angle 5

Mathematics

2 answers:

Daniel [21]3 years ago

5 0

Answer:

Angle 5 and angle 6 are supplementary angles

Step-by-step explanation:

Angle 5 and 6 are both supplementary angles because angle 6 and 5 are linear pairs

N76 [4]3 years ago

4 0

1 is supplementary to angle 5

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What does M equal in 4-m/2=10

Verizon [17]

Simplify both sides of your equation.

Subtract 4 from both sides.

Multiply both sides by 2(-1).

m = -12

3 0

2 years ago

Which equation is represented by the table

lyudmila [28]

Answer:

B. b = 3a + 2

Step-by-step explanation:

We can write the equation in slope-intercept form as b = ma + c, where,

m = slope/rate of change

c = y-intercept/initial value

✔️Find m using any two given pair of values, say (2, 8) and (4, 14):

Rate of change (m) = change in b/change in a

m = (14 - 8)/(4 - 2)

m = 6/2

m = 3

✔️Find c by substituting (a, b) = (2, 8) and m = 3 into b = ma + c. Thus:

8 = 3(2) + c

8 = 6 + c

8 - 6 = c

2 = c

c = 2

✔️Write the equation by substituting m = 3 and c = 2 into b = ma + c. Thus:

b = 3a + 2

4 0

2 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$