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Eduardwww [97]
3 years ago
12

Write the coordinate point for the vertex of this parabola: x=1/4y^2

Mathematics
1 answer:
ololo11 [35]3 years ago
5 0
Hello :
x =(1/4)y²
<span> the coordinate point for the vertex is :  O ( 0, 0)</span>
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What is the range of the relation[(-5,0),(-4,-3),(-3,4),(-1,0),(-4,-1)]
vesna_86 [32]

The range is the set of values that the relation takes for the domain for which it is defined.

Is the set of values of the dependent variable.

In this case, we have the relation [(-5,0),(-4,-3),(-3,4),(-1,0),(-4,-1)]​.

The dependant variable takes the values: 0, -3, 4, 0 and -1. Some values are repeated.

We put the repeated values only once and sort to write the range.

Then, the range is R = {-3, -1, 0, 4}.

Answer: Range = {-3, -1, 0, 4}

4 0
1 year ago
A<br> 9<br> Kylie drives 291 miles in 6 hours.<br> (a) What is her average speed?
ivolga24 [154]

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48.5 m/s

Step-by-step explanation:

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2 years ago
Three less than x is
pantera1 [17]

Answer:

x - 3

Step-by-step explanation:

3 0
2 years ago
Which number can each term of the equation be multiplied by to eliminate the decimals before solving?
Sati [7]
Ok...

5.6 = 56/10 = 560/100

1.1 = 11/10 = 110/100

0.12 = 12/100

--------------

\frac { 560 }{ 100 } j-\frac { 12 }{ 100 } =4+\frac { 110 }{ 100 } j\\ \\ \\ 100\times \left( \frac { 560 }{ 100 } j-\frac { 12 }{ 100 }  \right) =\left( 4+\frac { 110 }{ 100 } j \right) \times 100\\ \\ 560j-12=400+110j\\ \\ 560j-110j=400+12\\ \\ 450j=412\\ \\ j=\frac { 412 }{ 450 }

So, the answer is: 100

You could multiply both sides of the equation by 100 to get the value of (j) quickly.
6 0
3 years ago
Read 2 more answers
To one-one functions g and h are defined as follows.
Andreyy89

Answer:

  • 9
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Step-by-step explanation:

The inverse function for a set of ordered pairs can be found by swapping the x- and y-coordinates in each pair.

  g^{-1}(-1)=9\qquad\text{from the $g(x)$ ordered pair $(9, -1)$}

__

The inverse of a function expressed algebraically can be found by swapping the x- and y-variables and solving for y.

  h^{-1}(x)\qquad\text{is found from }x=h(y)\\\\x=3y+8\\x-8=3y\\\\y=\dfrac{x-8}{3}\\\\\boxed{h^{-1}(x)=\dfrac{x-8}{3}}

A function of its own inverse returns the original value:

  \boxed{\left(h\circ h^{-1}\right)(-1)=-1}

6 0
3 years ago
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