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2 years ago

Write and simplify an expression to model the relationship expressed in the situation below.

Mathematics

1 answer:

spayn [35]2 years ago

6 0

V(p) = x-n, where V(p) is the volume after the boxes have been together, x is the volume of the larger box, and n is the volume of the smaller box.
x = 15 cm x 25 cm x 20 cm = 7500 cubic centimeters (cm^3)
n = 10 cm x 10 cm x 10 cm = 1000 cm^3
V(p) = 7500 - 1000 = 6500 cm^3
So your answer is 6500 cubic centimeters.

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Which of the following fractions is equivalent to (−13)÷(−6)? Select two that apply.

tangare [24]

Answer:

D and F

Step-by-step explanation:

Please mark brainliest.

5 0

2 years ago

A ruby throated hummingbird beat its wings about 53 times each second about how many times does a ruby throated hummingbird beat

Sergeeva-Olga [200]

You would cross multiply 53 times 5 and divide by 1

3 0

2 years ago

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

$\to x$ )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence $a_n$ such that:

$a_n\ belongs\ to\ Q$

and

$a_n\to \alpha$

$f(a_n)=ka_n$

( since $a_n$ belongs to Q )

Let f is continuous at x=α

This means that:

$f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha$

This means that:

$f(\alpha)=k\alpha$

This means that:

f(x)=kx for every x∈ R

4 0

2 years ago

What is 1/2 times 6?

son4ous [18]

1/2 times 6 = 1/2 * 6
so in order to make this multiplication you have to turn 6 to a fraction, 6 is equal to 6/1
so , the way to multiply them is multiply the top with the top and it will be the top in the answer and the exact same with the bottoms

6*1= 6
2*1=2

so the fraction is 6/2, since a fraction is basicly a division

6/2 = 3

so the answer to your question is 3

i hope i helped you

6 0

2 years ago

Read 2 more answers

What are the solutions to the following system of equations? y=x^2+12x+30 and 8x-y=10

Ierofanga [76]

Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:

8x-10 = x^2+12x+30

Group all terms together on the right side. To do this, add -8x+10 to both sides. Then 0 = x^2 +4x +40. You must now solve this quadratic equation for x, if possible. I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.

If you have a graphing calculator, please graph 8x-10 and x^2+12x+30 on the same screen. You will see two separate graphs that do NOT intersect. This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.

6 0

3 years ago