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sammy [17]
3 years ago
13

Fido and Jet are two medium size dogs. Fido weighs exactly 15 pounds​ less than Jet. Together, they weigh exactly 89 pounds.​

Mathematics
1 answer:
Elan Coil [88]3 years ago
3 0

Answer:

jet weighs 74 pound

Step-by-step explanation:

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#10-1 : What is the probability of randomly choosing a diamond card from a regular deck of 52 cards? What is the probability of
Alchen [17]

\huge{\boxed{\frac{1}{4}}}\ \ \huge{\boxed{\frac{1}{4}}}

There are 13 of each suit in a deck of 52 cards.

This means the probability of drawing a diamond card is \frac{13}{52}. You can divide the numerator and denominator each by 13 to simplify. \frac{13}{52} \div \frac{13}{13} = \boxed{\frac{1}{4}}

Since there are the same number of clubs as there are diamonds, the probabilities are the same.

3 0
3 years ago
Help help pls help quick pls help help
uysha [10]
Answer: I’m pretty sure its D) 1 1/9
5 0
3 years ago
Read 2 more answers
4k+mn=n-3; solve for n
solong [7]
Since we are solving for n you have to isolate the n. Therefore, you want all the variables with an n in it on one side and all the variables without an n on the other side:) :
4k+3= -mn+n (There were a change in signs because for example you moved the -3 to the other side, therefore it can only be a negative on one side so now you have to change the sign to a positive:) hope you got that)
That would've been your answer because you cannot do anything else...Hope this helped :)

8 0
3 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
3 years ago
Please help if you can:)
Minchanka [31]
EXPLANATION:
Let the no. of orders Greg served be x.
Orders Chau served = 2x
Orders Maya served= x + 9
Orders served in all= 69

x + 2x + x + 9 = 69
=> 4x + 9 = 69
=> 4x = 69 - 9
=> x = 60/4
=> x = 15

ANSWER:
Number of orders Maya served = (15 + 9) 24
Number of orders Greg served = 15
Number of orders Chau served = (2*15) 30

Hope it helps u!
3 0
3 years ago
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