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erastovalidia [21]
3 years ago
9

Reflect (4,3) across the x-axis. Then reflect the result across the y-axis. What are the coordinates of the final point​

Mathematics
1 answer:
Tatiana [17]3 years ago
7 0

Answer: (-4,-3)

Step-by-step explanation:

(4,3) quadrant 1

(-4,3)quadrant 2 wasn't used in this

(4,-3) quadrant 3

(-4,-3)Quadrant 4

just trust me and I don't know how this is

college homework lol but okay

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MrMuchimi

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3 years ago
How to find the circumference for a 3 inch circle using 3.14 for pi
Vladimir79 [104]

Answer: C is approximately 9.42.

Step-by-step explanation: the radius is 1.5 and the C formula is 2 x pi x r = 2 (3.14)(1.5)

6 0
3 years ago
HELPP I NEED TO TURN THIS IN 5 Mins!!!!!
zaharov [31]

Answer:

3. The vertex form of the function, f(x) = x² - 4·x - 17 is f(x) = (x - 2)² - 21

4. The solutions are, x = -2 + √10 and x = -2 - √10

5. The quadratic equation with vertex (3, 1) and a = 1 in standard form is given as follows;

f(x) = x² - 6·x + 10

Step-by-step explanation:

3. The function given in standard form is f(x) = x² - 4·x - 17, which is the form, f(x) = a·x² + b·x + c

The vertex form of the of a quadratic function can be presented based on the above standard form as follows;

f(x) = a(x - h)² + k

Where;

(h, k) = The coordinate of the vertex

h = -b/2a

k = f(h)

Comparing with the given equation, we have;

f(x) = a·x² + b·x + c = x² - 4·x - 17

a = 1

b = -4

c = -17

∴ h = -(-4)/(2 × 1) = 2

h = 2

k = f(h) = f(2) = 2² - 4 × 2 - 17 = -21

k = -21

The vertex form of the function, f(x) = x² - 4·x - 17 is therefore, given as follows;

f(x) = (x - 2)² - 21

4. The given equation for which we need to solve by completing the square is 2·x² + 8·x = 12

Dividing the given equation by 2 gives;

x² + 4·x = 6

Which is of the form, x² + b·x = c

Where;

a = 1

b = 4

c = 6

From which we add (b/2)² to both sides to get x² + b·x + (b/2)² = c + (b/2)²

Adding (b/2)² = (4/2)² to both sides of x² + 4·x = 6 gives;

x² + 4·x + 4 = 6 + 4

(x + 2)² = 10

x + 2 = ±√10

x = -2 ± √10

The solution are, x = -2 + √10 and x = -2 - √10

5. Given that the value of the vertex = (3, 1), and a = 1, we have;

The vertex, (h, k) = (3, 1)

h = 3, k = 1

Therefore, h = 3 = -b/(2 × a) = -b/(2 × 1)

∴ -b = 2 × 3 = 6

b = -6

k = f(h) = a·h² + b·h + c, by substitution, we have;

k = f(3) = 1 × 3² + (-6) × 3 + c = 1

∴ c = 1 - (1 × 3² + (-6) × 3) = 10

c = 10

The quadratic equation with vertex (3, 1) and a = 1 in standard form, f(x) a·x² + b·x + c is therefor;

f(x) = x² - 6·x + 10

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Crm%5Cfrac%7Bd%7D%7Bdx%7D%20%20%20%5Cleft%20%28%20%5Cbigg%28%20%5Cint_%7B1%7D%5E%7B%20%7B
Margaret [11]

Applying the product rule gives

\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}

Use the fundamental theorem of calculus to compute the remaining derivatives.

\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt

The remaining integrals are

\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}

\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)

and so the overall derivative is

\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)

which could be simplified further.

5 0
2 years ago
Please help me with this one.
LUCKY_DIMON [66]

Answer: AC/EA = AB/DA

Step-by-step explanation:

7 0
3 years ago
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