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chubhunter [2.5K]
2 years ago
9

Concept Check: Sketch each angle in standard position. Draw an arrow representing the correct amount of rotation. Find the measu

re of two other angles, one positive and one negative, that are coterminal with the given angle. Give the quadrant of each angle, if applicable.
75°

174°

300°

-61°

-90°
Mathematics
1 answer:
soldi70 [24.7K]2 years ago
8 0

Step-by-step explanation:

...............

......

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prohojiy [21]
It will take 2 years.

Year 1 :
4% 2.5k = 2.5k/100 × 4 = $100 in interest

$2.5k + $0.1k = $2.6k

Year 2 :
4% 2.6k = 2.6k/100 × 4 = $104 in interest

$100 + $104 = $204 > $200

There may be a simpler way to solve this using a formula for example, but I just solved this step by step. Please feel free to ask if you have any questions
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For 3. he answer is a
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What is -18+6+3r+13r=
kvasek [131]
If this is just an expression, then all you have to do is combine like terms, or groups of numbers and variables that have the same variable.

When simplified, you get the equation

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Does anybody have any tips for calculating slopes in Algebra 1???
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Answer:

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Step-by-step explanation:

6 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
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