All categories

3 years ago

Is the image an example of an angle?

Mathematics

2 answers:

tatiyna3 years ago

8 0

Answer:

Yes; AB and BC do not form a line and share an endpoint.

Step-by-step explanation:

I just took the test and this is the correct answer.

andrew11 [14]3 years ago

7 0

Answer:

AC. ____ 2. What is another name for line m? a. line JG c. DB b. JGB. → ... Are points B, A, D, and C coplanar? Explain. a. No; one is on plane P. b. Yes ... False; the points to not have to form right angles. d. .... Name all segments parallel to AB. a.

Step-by-step explanation:

You might be interested in

Find the slope given (2,7) and (4,4)

8090 [49]

<h3>Answer:</h3>

m= -3/2

m=y1-y2/x1-x2 = 4-7/4-2 = -3/2

8 0

3 years ago

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.

4 0

3 years ago

Please help me with this homework

soldi70 [24.7K]

Step-by-step explanation:

we see f(x), which is |x|.

and we see g(x), which is clearly the same basic graph, it is just shifted 3 units down in y direction.

shifts are simply done by keeping the original function definition and then add it subtract a certain constant that then adapts every original functional value.

so, a shift down by 3 units is done by adding -3.

therefore, C is the correct answer (the original f(x) - 3).

8 0

2 years ago

Add the following polynomials:<br> 3a3 + 8a - 6 and 4a2 - 9a + 11

liubo4ka [24]

3 0

3 years ago

Read 2 more answers

Find the value of x.

mylen [45]

Answer: