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3 years ago

I need to find the measure of angle

Mathematics

1 answer:

lesantik [10]3 years ago

3 0

The answer is 124 deg. because all the equation is 3x+40=7x-72 and when you solve for x, you get 28. Then all you have to do is plug 28 into the equation 3x+40 and you get 124..:)):

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In need of some help please.

Hatshy [7]

By the Pythagorean theorem
50^2 = x^2 +(x +34)^2
2500 = 2x^2 +68x + 1156
x^2 +34x -672 = 0
(x -14)(x +48) = 0
x = 14 or -48

The distance from the wall to the base of the ladder is 14 ft.

_____
7-24-25 is a Pythagorean triple. The dimensions here are double those values. It can be handy to know a few of the Pythagorean triples, as that can let you write down the answers to problems without having to go through the equations.

8 0

3 years ago

I am stuck need to know the answer

just olya [345]

   OBAN
   92
|
|    STANRAER
172-------------------------------------112

To read this table you have to draw a vertical from OBAN downward aa horizontal line from the 1st value of STANRAER. The intersection point show the number of miles.
So there are 172 miles & the consumption is 8 mi/gal,. He needs 172/8 gal.
Cost number of gallons needed x price/ gallon
COST = (172/8) x 0.83
COST = (21.5) x 0.83 = 17.845 ≈ $17.85 (1st answer)

5 0

3 years ago

Simplify the expression in detail plz

gayaneshka [121]

(18t^2 + 9t^2) + (−7t −3t) + 20

27t^2 − 10t + 20

5 0

3 years ago

Any 10th grader solve it for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$