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2 years ago

Solve 4x + 11 = k for x. O A. x= -11 O B. x=-11 O C. x = 4k - 44 D. x= k-7

Mathematics

1 answer:

Wewaii [24]2 years ago

6 0

Answer:

4x + 11 = k

<=> 4x = k - 11

<=>

$x = \frac{k - 11}{4}$

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-6 (-2/3)<br> How is it solved

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You multiply -6 and -2/3. Which is 4

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2 years ago

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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.

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2 years ago

Pls help!!! Which steps would you use to solve this equation? Check all that apply.

hram777 [196]

Answer:

you subtract 7 from both sides

Step-by-step explanation:

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2 years ago

PQ is bisected at point R by MN. Which of the following I true about point R.

Stella [2.4K]

Answer:

C. R is the midpoint of PQ.

Step-by-step explanation:

Given that segment PQ is bisected at point R by MN.

It means PQ is divided into two parts at point R by the segment MN.

That is, PR = RQ

Hence, R is the midpoint of PQ.

It is not given that the segment MN is bisected by PQ.

So, R need not be the midpoint of MN.

Please refer to the attached figure for better understanding.

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3 years ago

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Iteru [2.4K]

Answer:

a. 45 π

b. 12 π

c. 16 π

Step-by-step explanation:

If a 3×5 rectangle is revolved about one of its sides of length 5 to create a solid of revolution, we can see a cilinder with:

Radius: 3

Height: 5

Then the volume of the cylinder is:

V=π*r^{2} *h= π*(3)^{2} *(5) = π*(9)*(5)=45 π

b. If a 3-4-5 right triangle is revolved about a leg of length 4 to create a solid of revolution. We can see a cone with:

Radius: 3

Height: 4

Then the volume of the cone is:

V=(1/3)*π*r^{2} *h= (1/3)*π*(3)^{2} *(4) = (1/3)*π*(9)*(4)=12 π

c. We can answer this item using the past (b. item) and solving for the other leg revolution (3):

Then we will have:

Radius: 4

Height: 3

Then the volume of the cone is:

V=(1/3)*π*r^{2} *h= (1/3)*π*(4)^{2} *(3) = (1/3)*π*(16)*(3)=16 π

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2 years ago