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3 years ago

A quadratic equation is shown below:

Mathematics

1 answer:

Mariulka [41]3 years ago

7 0

hello :

help :
the discriminat of each quadratic equation : ax²+bx+c=0 ....(a ≠ 0) is :
Δ = b² -4ac
1 ) Δ > 0 the equation has two reals solutions : x = (-b±√Δ)/2a
2 ) Δ = 0 : one solution : x = -b/2a
3 ) Δ < 0 : no reals solutions

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A diver takes a running leap off of a 40 foot cliff into a pool of water. When she is 5 feet from the cliff,

padilas [110]

of" (and any subsequent words) was ignored because we limit queries to 32 words.

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3 years ago

Find the square. (1/4A + 1/4B)^2 a) 1/16A^2 + 1/8AB + 1/16B^2 b) 1/4A^2 + 1/8AB + 1/4B^2

alisha [4.7K]

$(a + b)^2 = a^2 + 2ab + b^2$

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B^2$

Answer: A.

Another way to solve by factoring 1/4.

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= [\dfrac{1}{4}(A + B)]^2$

$= (\dfrac{1}{4})^2(A + B)^2$

$= \dfrac{1}{16}(A^2 + 2AB + B^2)$

$= \dfrac{1}{16}A^2 + \dfrac{1}{16}2AB + \dfrac{1}{16}B)^2$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B)^2$

6 0

3 years ago

Read 2 more answers

A projector displays an image on a wall. the area(in square feet) of the rectangular projection can be represented by

andrew-mc [135]

Answer:

The height of the projection will be (x - 5) feet.

Step-by-step explanation:

A projector displays an image on a wall. The area (in square feet) of the rectangular projection can be represented by (x² - 8x + 15).

Now to get the width and the height of the rectangular projection we have to factorize the expression (x² - 8x + 15).

Here, (x² - 8x + 15)

= x² - 3x - 5x + 15

= (x- 3)(x - 5)

Now, if the height of the projection is less than its width then, the height of the projection will be (x - 5) feet. (Answer)

3 0

3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

7 0

2 years ago

At the start of 2014, Mike's car is worth 12000. the car depreciates by 30 percent every year. how much is his car worth in 2017

Anarel [89]

Answer:

$4,116

Step-by-step explanation:

Worth of Mike's car at the start of 2014 = $12,000

If the car is said to depreciates every year by 30% = 30/100 = 0.3

The worth of the car at the start of 2017 is what we are to determine.

This means that the car depreciated by 30% (0.3) for 3 years since 2014 (2017 - 2014 = 3 yrs)

The worth at the start of 2017 would be calculated as follows:

12,000 × (1 - 0.3)³

= 12,000 × (0.7)³

= 12,000 × 0.343

= 4,116

Worth of the car at the start of 2017 would be $4,116

4 0

3 years ago