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3 years ago

The difference of two numbers is 9 and the sum of the numbers is 55. what are the numbers?

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

3 0

The two numbers I think is 3&6 because this is a trick question

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What is the difference of the polynomials?

Vilka [71]

Answer:

$8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6$

Step-by-step explanation:

We want to simplify;

$(8r^6s^3-9r^5s^4+3r^4s^5)-(2r^4s^5-5r^3s^6-4r^5s^4)$

We expand the bracket to obtain;

$8r^6s^3-9r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6+4r^5s^4$

We now group the like terms to obtain;

$8r^6s^3-9r^5s^4+4r^5s^4+3r^4s^5-2r^4s^5+5r^3s^6$

We now simplify to get;

$8r^6s^3-5r^5s^4+r^4s^5+5r^3s^6$

The correct answer is C

3 0

3 years ago

Read 2 more answers

1000000 x 0.3=? ready

svlad2 [7]

Answer:

300000

Step-by-step explanation:

Three hundred thousand

3 0

3 years ago

100 POINTS, WILL MARK BRAINLIEST! If h(x) = f[f(x)] use the table of values for f and f ′ to find the value of h ′(1).

victus00 [196]

Answer:

The value of $h^\prime(1)=5$

Step-by-step explanation:

Given that $h(x)=f(f(x))$

now to find $h(x)=f(f(x))$ from the given functions f(x) and f'x

let $h(x)=f(f(x))$

Then put x=1 in above function we get

$h(1)=f(f(1))$

$=f(3)$ (from the table f(1)=3 and f(3)=6)

Therefore h(1)=6

Now to find h'(1)

Let

$h^{\prime}(x)=f^{\prime}(f^\prime(x))$ (since $h(x)=f(f(x))$ )

put x=1 in above function we get

$h^{\prime}(1)=f^{\prime}(f^\prime(1))$

$=f^{\prime}(2)$ (From the table $f^\prime(1)=2$ and $f^\prime(2)=5$ )

$h^{\prime}(1)=5$

Therefore $h^{\prime}(1)=5$

4 0

3 years ago

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I have 1 apple I need 6 more to get to 7 so I pick 3 up eat 2 which picking up 4 more , how many apples do I have?

natita [175]

Answer:

Step-by-step explanation:

1+3=4 4-2=2 2+4=6

7 0

3 years ago

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What's the answer ???????

Tatiana [17]

A²+b²=c²
64+b²=196
b²=132
b=11.48912529

8 0

3 years ago