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Leona [35]
3 years ago
15

Find the value of x when r// s

Mathematics
1 answer:
ASHA 777 [7]3 years ago
5 0

Answer:

x = 81°

Step-by-step explanation:

Refer to the attachment for the steps.~

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3. Solve each equation, and check your solution.
Alisiya [41]

1) 2m - 16 = 2m + 4

2m - 2m = 4 + 16

0 = 20 (no solution)

2) -4(r + 2) = 4(2 - 2r)

-(r + 2) = 2 - 2r

-r - 2 = 2 - 2r

-r + 2r = 2 + 2

r = 4

3) 12(5 + 2y) = 4y - (6 - 9y)

60 + 24y = 4y - 6 + 9y

60 + 24y = 13y - 6

24y - 13y = -6 - 60

11y = -66

y = -6

7 0
3 years ago
What are the potential solutions to the equation 2ln(x+3)=0
skad [1K]
2\ln(x+3)=0\implies \ln(x+3)^2=0\implies e^{\ln(x+3)^2}=e^0\implies (x+3)^2=1

Expanding the left side gives

x^2+6x+9=1\implies x^2+6x+8=0\implies (x+4)(x+2)=0

which gives two solutions, x=-4 and x=-2. But if x=-4, then \ln(x+3)=\ln(-1), but this number isn't real, so x=-4 is an extraneous solution. Meanwhile if x=-2, you get \ln(-2+3)=\ln1=0, so this solution is correct.

"Potential solutions" might refer to both possibilities, but there is only one actual (real) solution.
4 0
3 years ago
Read 2 more answers
What is the volume of a rectangular prism with the following dimensions
andriy [413]

Answer: 250ft3

Step-by-step explanation:

v=l*w*H

v= 10*5 * 5

v=250

7 0
3 years ago
Read 2 more answers
Find the length of B to the nearest tenth using the Pythagorean theorem.​
bekas [8.4K]

Step-by-step explanation:

Here

BC=P=10

AC=B=b

AB=H=15

then using fourmula

h^2= p^2+ b^2

b^2= 15×15 - 10× 10

b^2= 125

b=11.2

4 0
3 years ago
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; (x^2-4)^2y'' + (x + 2)y' + 7y = 0

The above equation can be better expressed as:

y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

p(x) = \dfrac{(x+2)}{(x^2-4)^2} \

p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \

p(x) = \dfrac{1}{(x+2)(x-2)^2}

Also;

q(x) = \dfrac{7}{(x^2-4)^2}

q(x) = \dfrac{7}{(x+2)^2(x-2)^2}

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

\lim \limits_{x \to-2} (x+ 2) p(x) =  \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{1}{(x-2)^2}

\implies \dfrac{1}{16}

\lim \limits_{x \to-2} (x+ 2)^2 q(x) =  \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{7}{(x-2)^2}

\implies \dfrac{7}{16}

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0
3 years ago
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