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3 years ago

What are the roots of f(x)=x2-48

Mathematics

2 answers:

Tcecarenko [31]3 years ago

5 0

I do believe that would be x=6 and x=-8

olga_2 [115]3 years ago

3 0

The roots are 4(squared sign) 3 and -4(squared sign) 3

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How many integers are there that consist of four digits (from 1000 to 9999) where the first digit is a 5 or an 8 and the third d

Ostrovityanka [42]

Answer:

1600 integers

Step-by-step explanation:

Since we have a four digit number, there are four digit placements.

For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.

For the second digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.

For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers

5 0

3 years ago

Whats the area of a rectangle 17ft 4ft

Katena32 [7]

Answer:

Step-by-step explanation:

Multiply the numbers to get the area

7 0

3 years ago

Read 2 more answers

Rewrite the following expression.

faust18 [17]

We have been given the expression

$x^{\frac{9}{7}}$

We have the exponent rule

$x^{\frac{a}{b}}= x^{a^{(\frac{1}{b})}}$

Using this rule, we have

$x^{\frac{9}{7}}= x^{9^{(\frac{1}{7})}}$

Now, using the fact that $x^{\frac{1}{n}}=\sqrt[n]{x}$ , we get

$x^{\frac{9}{7}}= \sqrt[7]{x^9}\\
\\
x^{\frac{9}{7}}=\sqrt[7]{x^7\times x^2}\\
\\
x^{\frac{9}{7}}=x\sqrt[7]{x^2}$

D is the correct option.

6 0

3 years ago

Read 2 more answers

Eddi Din [679]

Answer:

a) $y=\dfrac{5}{2}x$

b) yes the two lines are perpendicular

c) $y=\dfrac{5}{4}x+6$

Step-by-step explanation:

a) All this is asking if to find a line that is perpendicular to 2x + 5y = 7 AND passes through the origin.

so first we'll find the gradient(or slope) of 2x + 5y = 7, this can be done by simply rearranging this equation to the form $y = mx + c$

$5y = 7 - 2x$

$y = \dfrac{7 - 2x}{5}$

$y = \dfrac{7}{5} - \dfrac{2}{5}x$

$y = -\dfrac{2}{5}x+\dfrac{7}{5}$

this is changed into the $y = mx + c$ , and we easily see that -2/5 is in the place of m, hence $m = \frac{-2}{5}$ is the slope of the line 2x + 5y = 7.

Now, we need to find the slope of its perpendicular. We'll use:

$m_1m_2=-1$ .

here both slopes $m_1$ and $m_2$ are slopes that are perpendicular to each other, so by plugging the value -2/5 we'll find its perpendicular!

$\dfrac{-2}{5}m_2=-1$ .

$m_2=\dfrac{5}{2}$ .

Finally, we can find the equation of the line of the perpendicular using:

$(y-y_1)=m(x-x_1)$

we know that the line passes through origin(0,0) and its slope is 5/2

$(y-0)=\dfrac{5}{2}(x-0)$

$y=\dfrac{5}{2}x$ is the equation of the the line!

b) For this we need to find the slopes of both lines and see whether their product equals -1?

mathematically, we need to see whether $m_1m_2=-1$ ?

the slopes can be easily found through rearranging both equations to $y=mx+c$

Line:1

$2x + 3y =6$

$y =\dfrac{-2x+6}{3}$

$y =\dfrac{-2}{3}x+2$

Line:2

$y = \dfrac{3}{2}x + 4$

this equation is already in the form we need.

the slopes of both equations are

$m_1 = \dfrac{-2}{3}$ and $m_2 = \dfrac{3}{2}$

using

$m_1m_2=-1$

$\dfrac{-2}{3} \times \dfrac{3}{2}=-1$

$-1=-1$

since the product does equal -1, the two lines are indeed perpendicular!

c)if two perpendicular lines have the same intercept, that also means that the two lines meet at that intercept.

we can easily find the slope of the given line, y = − 4 / 5 x + 6 to be $m=\dfrac{-4}{5}$ and the y-intercept is $c=6$ the coordinate at the y-intercept will be (0,6) since this point only lies in the y-axis.