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2 years ago

HELPPPP ASAAPPPP PLEASEEEE

Mathematics

1 answer:

lidiya [134]2 years ago

6 0

(A) n=w by the corresponding angles theorem.

(B) This is the correct answer. n=s by the vertical angles theorem and s and x are supplementary by the same-side interior angles theorem. Thus, n and x are supplementary and not always equal.

(D) s=w, so (s+w)/2 = (s+s)/2 = s. Since n=s by the vertical angles theorem, n=s.

(E) n and w are equal by the corresponding angles theorem, and w and y are supplementary as they form a linear pair. Thus, n+y=180, and it follows n=180-y.

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9. Find mZDBC in the figure below.

ArbitrLikvidat [17]

Answer:

a. 66 degrees

Step-by-step explanation:

3 0

3 years ago

ILL GIVE BRAINLIST ON WHOEVER GETS RIGHT!!!!!!!

Korvikt [17]

Answer:

hey,its option a

Step-by-step explanation:

because angle F= angle M and angle G= angle L

and angle H= angle N

To know third angle in ∆HGF

,we know sum of all angles in a triangle is 180°

______________________________

let angle H be unknown value x

x+51+36=180

x=180-87 ---------------- 1

similarly in ∆MNL

let angle be y

51+36+y=180

y=180-87----------------2

_______________________________

From 1&2 we observe qngle x= angle y

by AAA SIMILARITY THEY ARE SIMILAR

3 0

4 years ago

Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$