All categories

3 years ago

Match the input and outputs for the relation equation image indicator

Mathematics

2 answers:

e-lub [12.9K]3 years ago

5 0

There is no image here

Cloud [144]3 years ago

4 0

Answer:
There is no image here
Explanation:

You might be interested in

Rewrite in simplest radical form 1/x^-3/6.

Aneli [31]

$\cfrac{1}{x^{- \frac{3}{6} }} = \cfrac{1}{x^{- \frac{1}{2} }} =x^{ \frac{1}{2} }= \sqrt{x}$

6 0

3 years ago

Y=5x-4

anyanavicka [17]

Answer:

$\displaystyle -5x + y = -92\:or\:y = 5x - 92$

Step-by-step explanation:

−12 = 5[16] + b

$\displaystyle -92 = b \\ \\ y = 5x - 92$

If you want it in Standard Form:

y = 5x - 92

- 5x - 5x

__________

$\displaystyle -5x + y = -92$

I am joyous to assist you anytime.

7 0

2 years ago

How to do it ? plz answer it..

Vera_Pavlovna [14]

I hope you understand....

6 0

3 years ago

Which equation can be solved to find one of the missing side lengths in the triangle?aB60512 unitsАCOS(60°) =22 믕cos(60°) = 12Su

Mekhanik [1.2K]

In a right rectangle, we have:

$\sin \alpha=\frac{opposite}{hypotenuse}$ $\cos \alpha=\frac{\text{adjacent}}{hypotenuse}$

For your exercice, hypotenuse=12

The exercise also inform the angle 60°, then:

$\sin \text{ 60}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{b}{12}$ $\cos 60=\frac{adjacent}{\text{hypotenuse}}=\frac{a}{12}$

3 0

1 year ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

5 0

3 years ago