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3 years ago

F ∠1 ≅ ∠3, which conclusion can be made?

Mathematics

1 answer:

vichka [17]3 years ago

4 0

Answer:

I think your answer would be A.

Step-by-step explanation:

If numer 1 and number ≅ then a and b would be parallell i think

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Solving systems of equations algebraically.

Elanso [62]

Answer:

x = 2

y = -1

Step-by-step explanation:

Since it gives you the value of y already, just plug that into the equation, like so:

y = 4x - 9

x - 3 = 4x - 9

Subtract the x from both sides:

x - 3 = 4x - 9

-x - x

__________

-3 = 3x - 9

Then, add 9 to both sides:

-3 = 3x - 9

+9 + 9

________

6 = 3x

Divide both sides by 3:

2 = x

Now plug in x for y:

y = x - 3

y = 2 - 3

y = -1

8 0

3 years ago

8(x + 30) + 5 A x + 35 B 8x + 35 C 8x + 240 D 8x + 245

Anon25 [30]

Answer:

Step-by-step explanation:

First, remove parenthesis:

8(x + 30) = 8*x + 8*30 = 8x + 240

Then, add the 5 at the end:

8x + 240 + 5 = 8x + 245

6 0

3 years ago

Read 2 more answers

if an amount of $5,000 is deposited into a savings account at an annual interest rate of 5%, compounded monthly, what is the val

Ugo [173]

Answer:

$8,240

Step-by-step explanation:

We are given that,

Principle amount in the savings account, P = $5,000.

Rate of interest, r = 5% = 0.05

Time period, t = 10

Also, the interest is compounded monthly, n = 12

As, we now that the value of the investment is given by $P(1+\frac{r}{n})^{nt}$

Thus, we have,

Investment Value = $5000(1+\frac{0.05}{12})^{12\times 10}$

i.e. Investment Value = $5000(1+\frac{0.05}{12})^{12\times 10}$

i.e. Investment Value = $5000(1+0.00417)^{120}$

i.e. Investment Value = $5000\times 1.648$

i.e. Investment Value = $8,240

Hence, the investment amount after 10 years is $8,240.

5 0

3 years ago

Julia charged $612 to her credit card. The cost of her next purchase caused her to go over her credit limit of $2000.

Fynjy0 [20]

She has already spent 612, and by adding C 612 has gone ABOVE 2000
So 612 + C MUST be Greater Than 2000
The answer is D

7 0

3 years ago

Read 2 more answers

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