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3 years ago

The fraction 5/6 is found between which pair of fractions on a number line

Mathematics

2 answers:

forsale [732]3 years ago

8 0

It is between 4/6 and 6/6

Dvinal [7]3 years ago

8 0

It is btween 5/5 and 5/7 easy.

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Help me to solve this question

Anastasy [175]

Answer:

$\frac{6}{25}$

Step-by-step explanation:

$P(A) = \frac{number \: of \: times \: A \: occurs}{total \: number \: of \: spins}$

$P(A) = \frac{12}{12 + 16 + 13 + 9} = \frac{12}{50} = \frac{6}{25}$

4 0

2 years ago

There are three questions for this problem <br> Please answer all three need help!!!

Vlad [161]

Yes what’s your question

8 0

4 years ago

Hall and Mindy are playing a guessing game. Hall tells Mindy: ""The difference between 17 and the square root of my mystery numb

cestrela7 [59]

Answer:

The possible number that Hall could be thinking of is 144.

Step-by-step explanation:

Difference between 17 and the square root of my mystery number

This is represented by:

$17 - \sqrt{x}$

is 5

Then:

$17 - \sqrt{x} = 5$

What are two possible numbers that Hall could be thinking of?

We have to solve the equation, for x. So

$17 - \sqrt{x} = 5$

$\sqrt{x} = 12$

$(\sqrt{x})^2 = (12)^2$

$x = 144$

The possible number that Hall could be thinking of is 144.

5 0

3 years ago

Write an equation of a line in slope-intercept form with the given slope and y-intercept.

Ganezh [65]

The equation is, y = 8x-3

5 0

2 years ago

Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

$-4=a(1-3)(1+1)$

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

3 0

4 years ago