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ikadub [295]
3 years ago
10

The television coverage of the soccer f.a. Final begin at 2:05 a.m. and finished an hour after 50 minutes later what was the tim

e when it finished
Mathematics
1 answer:
son4ous [18]3 years ago
8 0

Answer: 3:55 am is correct

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If f ( 4 ) = − 2 , write an ordered pair that must be on the graph of y = f ( x + 4 ) + 1
AlekseyPX

We have been given that

y = f ( x + 4 ) + 1\\
\\
f(4)=-2

Now, we need to find an ordered pair that must be on the graph of

y = f ( x + 4 ) + 1

On substituting x=0 in the above equation, we get

y = f ( 0 + 4 ) + 1\\
\\
y=f(4)+1\\
\\
\text{On substituting the value of f(4)= -2, we get}\\
\\
y=-2+1\\
y=-1

Therefore, the required ordered pair is given by (0,-1)

4 0
3 years ago
A retired math teacher takes his grandson shopping. He tells his grandson that they can spend x
leva [86]

Answer:

Addition, 83.6

Step-by-step explanation:

x - 72.6 = 11, x = 83.6

The grandson must add 72.6 to both sides to solve for x.

4 0
3 years ago
What’s 14/20 - 81/90
Leni [432]

Answer: -1/5

Step-by-step explanation:

14/20 - 81/90=

7/10-9/10=       ==> 10 is the GCF of 20 and 90

(7-9)/10=

-2/10=-1/5

3 0
1 year ago
Read 2 more answers
Plz help me with this!!!!!!!!
LenaWriter [7]

Answer:  √y  

<u>Step-by-step explanation:</u>

\sqrt[6]{y^3}=y^{\frac{3}{6}}=y^{\frac{1}{2}}=\boxed{\sqrt y}

4 0
3 years ago
What is the solution of x=2+ square root x-2?
alex41 [277]

So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

x-2=\sqrt{x-2}

Next, square both sides:

(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2

Next, subtract x and add 2 to both sides of the equation:

x^2-5x+6=0

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

x^2-2x-3x+6=0

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

x(x-2)-3(x-2)=0

Now you can rewrite the equation as (x-3)(x-2)=0

Now, apply the Zero Product Property and solve for x as such:

x-3=0\\x=3\\\\x-2=0\\x=2

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}

Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>

8 0
3 years ago
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