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2 years ago

12

Using the graphing function on your calculator, what is the solution to the system of equations show below?

1 answer:

horsena [70]2 years ago

4 0

Answer:

ima redo my answer but

Step-by-step explanation:

For y +2x5 is that an times sign or wht cause i can't answer that unless thats cleared up first

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Calculate f(x) for the given domain values.

Maksim231197 [3]

Answer:

See explanation

Step-by-step explanation:

1. The given function is

$f(x) = - 3x$

The domain values are: x=0, 2, -1, 4, -2

When x=0

$f(0) = - 3 \times 0 = 0$

When x=2,

$f(x) = - 3 \times 2 = - 6$

When x=-1

$f(x) = - 3 \times - 1 = 3$

When x=4

$f(4) = - 3 \times 4 = - 12$

When x=-2

$f( - 2) = - 3 \times - 2 = 6$

2. The given function is

$f(x) = \frac{1}{3} x$

When x=3,

$f(3) = \frac{1}{3} \times 3 = 1$

Similarly,

$f( - 3) = \frac{1}{3} \times - 3 = - 1$

$f(300) = \frac{1}{3} \times 300 = 100$

$f( - 180) = \frac{1}{3} \times - 180 = - 60$

$f(99) = \frac{1}{3} \times 99 = 33$

8 0

2 years ago

If q→ p is true, then ~ p → ~ q is _____ true. always sometimes never

Bogdan [553]

Answer: always

Step-by-step explanation:

6 0

3 years ago

A ball is thrown into the air with an upward

inn [45]

To find the 'maximum height' we will need to take the derivative of h(t) = –16t² + 32t + 6 then set it equal to zero, then solve for t. this t will be the time at which the ball reaches it's maximum height.

5 0

3 years ago

A certain number added to its square is 30. Find the number

Alex Ar [27]

X+x²=30
x²+x-30=0
(x+6)(x-5)=0
x=-6 or x=5

5 0

3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

or

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago