What has a value between -2 and -3

Name a number that can be evenly divided by 6 and and by 7

Alona [7]

There are an infinite number of them. The smallest one is 42,
and every multiple of 42 is another one.

I don't remember the official correct way to solve this problem.
I can only show you my own way of doing it.

I take the bigger number, and I go through its multiples, one by one,
until I spot one that's also a multiple of the smaller one.

Example: 6 and 7.
Go through the multiples of 7:

7 ... no, not a multiple of 6
14 ... no, not a multiple of 6
21 ... no, not a multiple of 6
28 ... no, not a multiple of 6
35 ... no, not a multiple of 6
42 ... Yes ! A multiple of 6. yay !

7 0

2 years ago

Hassan is cooking some rice the recipe says he needs 450 grams of rice for 6 people.how many kilograms of rice would he need for

gregori [183]

Answer:

so if we take 450 an divide it by 6 you get 75. So take 75 times 18 and you get 1,350 so he would need 1,350 kilograms of rice for 18 people

Step-by-step explanation:

hope this helps :D

4 0

2 years ago

The solution to an any quality is giving in the same building notation as The solution to an inequality is giving in the set bui

Lostsunrise [7]

Options

$(A)\left(-\infty , \dfrac23\right]\\\\(B)\left(-\infty , \dfrac23\right) \\\\(C)(\frac23\right, \infty ) \\\\(D) [\frac23\right, \infty )$

Answer:

$(C)(\frac23\right, \infty )$

Step-by-step explanation:

Given the solution to an inequality

{x|x>2/3}

The solution set does not include $\dfrac23$ , therefore, it must be open at the left. Recall that we use a curvy bracket ( to denote openness at the left.

Since x is greater than $\dfrac23$ , the solution set contains all values of larger than $\dfrac23$ up till infinity. Since infinity is an arbitrarily large value, we also use an open bracket at the right.

Therefore, another way to represent the solution {x|x>2/3} is:

$(\frac23\right, \infty )$

The correct option is C.

8 0

3 years ago

ANSWER ASAP PLEASE AND SHOW YOUR WORK!!!!

IgorLugansk [536]

Answer:

f(-x)= -f(x) so, the given function $f(x)= x^3-2x$ is Odd

Step-by-step explanation:

We need to determine if the function $f(x)= x^3-2x$ is Even, Odd, or Neither.

Determining if a function f(x) is even or not we put x=-x and check

If f(-x)=f(x) the function is even

if f(-x)= -f(x) the function is odd

So, Putting x =-x and determining if the given function $f(x)= x^3-2x$ is Even, Odd, or Neither.

$f(x)= x^3-2x\\Put \ x=-x\\f(-x)=(-x)^3-2(-x)\\f(-x)=-x^3+2x\\f(-x)=-(x^3-2x)\\f(-x)=-f(x)$

As f(-x)= -f(x) so, the given function $f(x)= x^3-2x$ is Odd

5 0

2 years ago

PLEASE HELP!!! very confused

Leni [432]

Answer:

Step-by-step explanation:

The vertices lie on the x-axis, as is determined by their coordinates. This makes the center of this hyperbola (0, 0) because the center is directly between the vertices. The fact that the foci also lie on the x-axis tells us that this is the main axis. What this also tells us is which way the hyperbola "opens". This one opens to the left and the right as opposed to up and down. The standard form for this hyperbola is:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ and so far we have that h = 0 and k = 0.

By definition, a is the distance between the center and the vertices. So a = 5, and a-squared is 25. So we're getting there. Now here's the tricky part.

The expressions for the foci are (h-c, k) and (h+c, k). Since we know the foci lie at +/-13, we can use that to solve for c:

If h+c = 13 and h = 0, then

0 + c = 13 and c = 13.

We need that c value to help us find b:

$c^2=a^2+b^2$ and