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2 years ago

Just answer this pls i need it

Mathematics

1 answer:

Aleks [24]2 years ago

3 0

Answer:

J is the answer 4 $\frac{5}{6}$

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The exchange rate is: £1 = $1.54 How many £ would I get for $77 Please help now

tiny-mole [99]

Answer:

£50

Step-by-step explanation:

£1 = $1.54

£x = $77

Cross multiply.

1.54x = 77

x = 77/1.54

x = 50

You would get £50 for $77.

4 0

2 years ago

Read 2 more answers

NEED HELP ASAP!

Brrunno [24]

complette the square to get vertex form or y=a(x-h)^2+k
(h,k) is vertex
1. group x terms, so for y=ax^2+bx+c, do y=(ax^2+bx)+c 2, factor out the leading coefinet (constant in front of the x^2 term), basicallly factor out a 3. take 1/2 of the linear coefient (number in front of the x), and square it ,then add negative and positive of it inside parnthases 4. complete the squre and expand 

so
y=-1/4x^2+4x-19
group
y=(-1/4x^2+4x)-19
undistribute -1/4
y=-1/4(x^2-16x)-19
take 1/2 of -16 and squer it to get 64 then add neg and pos inside
y=-1/4(x^2-16x+64-64)-19
factorperfect square
y=-1/4((x-8)^2-64)-19
expand
y=-1/4(x-8)^2+16-19
y=-1/4(x-8)^2-3
vertex is (8,-3)

8 0

3 years ago

Solve the proportion 17/4 = y/6

Tanya [424]

Answer:

This is your answer ☺️☺️

3 0

2 years ago

Find the product of the complex number and it's conjugate. 9-5i

Dennis_Churaev [7]

Conjugate of 9-5i is 9+5i
product of those two: (9-5i)*(9+5i) = $9^{2} - (5i)^{2} =81-25* i^{2} =81-25*(-1)=81+25=106$

3 0

3 years ago

If S is countable and nonempty, prove their exist a surjection g: N --> S

aleksley [76]

Answer with Step-by-step explanation:

We are given S be any set which is countable and nonempty.

We have to prove that their exist a surjection g:N $\rightarrow S$

Surjection: It is also called onto function .When cardinality of domain set is greater than or equal to cardinality of range set then the function is onto

Cardinality of natural numbers set = $\chi_0$ ( Aleph naught)

There are two cases

1.S is finite nonempty set

2.S is countably infinite set

1.When S is finite set and nonempty set

Then cardinality of set S is any constant number which is less than the cardinality of set of natura number

Therefore, their exist a surjection from N to S.

2.When S is countably infinite set and cardinality with aleph naught

Then cardinality of set S is equal to cardinality of set of natural .Therefore, their exist a surjection from N to S.

Hence, proved

8 0

2 years ago