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3 years ago

Find the domain and range of the function: f(t)=sec[((pi)t)/4]

Mathematics

1 answer:

beks73 [17]3 years ago

3 0

The domain and range of the function f(t)=sec[((pi)t)/4] are the following:

<span>Domain: All the real numbers except t = 2 + 4k, where k is an integer. </span>

Range: (-∞, -1] U [1, ∞)

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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2 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

$9dp = c^2$

$c^2 = 9dp$

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

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2 years ago

How is 3 divided by 1.4 ...2.14? Working please!

musickatia [10]

Turn 1.4 into a whole number. Which makes 1.4 to 14. We know that 3 has a decimal point at the end. So move the decimal point of 3 to the right, making it 30. Now do 30/14. You will get...

30/14 = 2.14

GG ;)

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3 years ago

Convert the linear equation 8x − 2y = 12 into function form.

alex41 [277]

Answer:

y=4x-6

Step-by-step explanation:

8x-2y=12

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4x-6=y

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2 years ago