Ask question

Ask question

All categories

3 years ago

13

There is an ice cream & candy shop next door that is the

1 answer:

ella [17]3 years ago

5 0

Answer:

it's the same side, so it'll be 15/30

You might be interested in

A distance runner is running an 18-mile race. So far, he has run 43% of the race.

Nezavi [6.7K]

The answer would be Find 40% of 20 because you would round 43% down and 18 up. Hope this helps :)

8 0

3 years ago

Read 2 more answers

Find the slope of the line

alekssr [168]

Answer:

hhhhhh

Step-by-step explanation:

hhhhhhhhhhhhhhhhhhhhhhh

3 0

3 years ago

The population of rabbits in an area can be modeled with an exponential function. In the first count, there are 250 rabbits. One

hjlf

The answer will be definitely be A don’t listen

6 0

3 years ago

The equation 2x2 − 12x + 1 = 0 is being rewritten in vertex form. Fill in the missing step. Given 2x2 − 12x + 1 = 0 Step 1 2(x2

Bezzdna [24]

Answer:

Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

Convert to vertex form

step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

$-2(x-2)^{2}+5=0$

$-2(x-2)^{2}=-5$

$(x-2)^{2}=2.5$

square root both sides

$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

$x=2(+)1.58=3.58$

$x=2(-)1.58=0.42$

Part 5) we have

$f(x)=-x^{2}+50x-264$

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

so

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

$-x^{2}+50x-264=0$

$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

$-(x^{2}-50x+625)=-361$

$(x^{2}-50x+625)=361$

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

$x=25(-)19=6$

Part 6) we have

$-2x^{2}+28x+20$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

$-2(x^{2}-14x+49)+118$

Rewrite as perfect square

$-2(x-7)^{2}+118$

The vertex is the point (7,118)

therefore

The video game price that produces the highest weekly profit is x=$7

Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

$f(x)-6=-(x^{2}-8x+16)$

Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

The vertex is the point (4,6)

therefore

The maximum height of the puck is 4 feet.

Part 8) we have

$x^{2}+6x+5$

Convert to vertex form

Group terms

$(x^{2}+6x)+5$

Complete the square

$(x^{2}+6x+9)+5-9$

$(x^{2}+6x+9)-4$

Rewrite as perfect squares

$(x+3)^{2}-4$

Part 9) we have

$2x^{2}-4x-2=0$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

$2(x^{2}-2x)-2=0$

Complete the square

$2(x^{2}-2x+1)-2-2=0$

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

$2(x-1)^{2}=4$

The vertex is the point (1,-4)

Part 10) we have

$8x^{2}-64x+720$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

$8(x^{2}-8x)+720$

Complete the square

$8(x^{2}-8x+16)+720-128$

$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

$8(x-4)^{2}+592$

the vertex is the point (4,592)

The population has a minimum at x=4 years ( that is after 4 years since 1998 )

6 0

3 years ago

Name the ordered pairs for the vertices on the figure below

Elanso [62]

Answer:

where is the figure?

Step-by-step explanation:

4 0

3 years ago