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2 years ago

12

What an angel adjacent to < igf

2 answers:

dem82 [27]2 years ago

7 0

Answer:

Two angles are Adjacent when they have a common side and a common vertex (corner point) and don't overlap. Because: they have a common side (line CB)

Step-by-step explanation:

Semmy [17]2 years ago

5 0

Answer:

B.) FGD

Step-by-step explanation:

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Whats -6-10k if k=-1

defon

Answer:

The answer would be 4

Step-by-step explanation:

Calculate it

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2 years ago

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Under which angle conditions could a triangle exist? Check all that apply.

Komok [63]

<span>Triangles exist:
</span>3 acute angles<span> 2 acute angles, 1 right angle
</span><span>2 acute angles, 1 obtuse angle
</span>
Triangles DON'T exist for:
<span>1 acute angle, 1 right angle, 1 obtuse angle
</span><span>1 acute angle, 2 obtuse angles</span>

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3 years ago

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Determine if the following infinite series converges or diverges

Mandarinka [93]

Using limits, it is found that the infinite sequence converges, as the limit does not go to infinity.

<h3>How do we verify if a sequence converges of diverges?</h3>

Suppose an infinity sequence defined by:

$\sum_{k = 0}^{\infty} f(k)$

Then we have to calculate the following limit:

$\lim_{k \rightarrow \infty} f(k)$

If the <u>limit goes to infinity</u>, the sequence diverges, otherwise it converges.

In this problem, the function that defines the sequence is:

$f(k) = \frac{k^3}{k^4 + 10}$

Hence the limit is:

$\lim_{k \rightarrow \infty} f(k) = \lim_{k \rightarrow \infty} \frac{k^3}{k^4 + 10} = \lim_{k \rightarrow \infty} \frac{k^3}{k^4} = \lim_{k \rightarrow \infty} \frac{1}{k} = \frac{1}{\infty} = 0$

Hence, the infinite sequence converges, as the limit does not go to infinity.

More can be learned about convergent sequences at brainly.com/question/6635869

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6 0

1 year ago

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

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3 years ago

Solve the equations to find the number and type of solutions

luda_lava [24]

Answer:

This has one real solution, x=4

Step-by-step explanation:

8 - 4x = 0

Add 4x to each side

8 - 4x+4x = 0+4x

8 =4x

Divide each side by 4

8/4 = 4x/4

2 =x

This has one real solution, x=4

6 0

3 years ago

Read 2 more answers