All categories

2 years ago

Which is the solution to the inequality? y+15<13 A:y<-12 B:y>-12 C:y<18 D:y>18

Mathematics

1 answer:

Umnica [9.8K]2 years ago

8 0

The answer is A because anything less than a negative will be just a even bigger negative

You might be interested in

Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water (in

Bingel [31]

The diver will hit the water when the height is zero, so we will want to set our height function equal to zero
0 = -4t^2 + 11t + 3
we will then factor to find out value of t that make the function equal to zero
0 = (-4t - 1 ) ( t - 3 )
This means our function equals zero when t = -1/4 s and t = 3 seconds
Since time cannot be negative, our final solution is t = 3 seconds

4 0

2 years ago

Read 2 more answers

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

Which of the following is the graph of a cubic function?

Vera_Pavlovna [14]

Answer:

the one that looks like s but sideways

Step-by-step explanation:

You didn’t show the graph

4 0

3 years ago

HELP HELP HELP HELP 15 Points

stich3 [128]

W=2j, m=5w, (m+4)=6(j+4)

using w from first in second you get:

m=5(2j)=10j, m+4=6j+24

m=10j, m=6j+20, since m=m we can say

10j=6j+20

4j=20

j=5, and since m=10j

m=50

So the man is 50 years old

5 0

3 years ago

Read 2 more answers

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

6 0

3 years ago