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3 years ago

H(x)=2x-1,find h(-1)

Mathematics

2 answers:

valina [46]3 years ago

7 0

Answer:

-3

Step-by-step explanation:

gladu [14]3 years ago

6 0

Answer:

-3

Step-by-step explanation:

Substitute -1 for x in the expression 2x-1

2*-1-1

-2-1

-3

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Select the graph that would represent the best presentation of the solution set.<br> \absP > 3

m_a_m_a [10]

Answer:

Step-by-step explanation:

there's no graph selection attached. Can you upload it and then I can help?

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2 years ago

Find intervals when the graph is growing or increasing or constant

AleksAgata [21]

The function shown in the graph is increasing over the interval (-1, 0) and (1, ∞). It is decreasing over the interval (-∞, -1) and (0, 1). The function does not remain constant.

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The function shown in the graph is increasing over the interval (-1, 0) and (1, ∞). It is decreasing over the interval (-∞, -1) and (0, 1). The function does not remain constant.

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2 years ago

Which are the solutions of x2 = –7x – 8?

vovangra [49]

Answer:

The solutions are:

$x_1=\frac{-7+\sqrt{17}}{2}$ $x_2=\frac{-7-\sqrt{17}}{2}$

Step-by-step explanation:

We have the following quadratic equation

$x^2 = -7x - 8$

We can rewrite the equation as follows

$x^2+7x + 8=0$

Now we use the quadratic formula to solve the equation

For an equation of the form $ax ^ 2 + bx + c = 0$ the quadratic formula is:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

In this case:

$a=1,\ b=7,\ c=8$

Then:

$x=\frac{-7\±\sqrt{7^2-4(1)(8)}}{2(1)}$

$x=\frac{-7\±\sqrt{49-32}}{2}$

$x=\frac{-7\±\sqrt{17}}{2}$

$x_1=\frac{-7+\sqrt{17}}{2}$

$x_2=\frac{-7-\sqrt{17}}{2}$

7 0

4 years ago

Read 2 more answers

Xavier ran four miles on his first day of training. The next day he ran eight ninths that distance. How far did he run on the se

serg [7]

Well, it's the product of 8/9 and the 4 miles. The result is 32/9 miles, that is, 3,55 miles.

7 0

3 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

3 years ago