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Keith_Richards [23]

3 years ago

8

There are 2 pecks in one kenning there are 2 kennings in 1 bushel there are 8 bushels in1 quarter there are 5 quarters in 1 load

write a number sentence to show how many pecks are in1 load

1 answer:

Delicious77 [7]3 years ago

7 0

Answer:

yes 1 2 3

Step-by-step explanation:

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Solve for Y MUST SHOW WORK<br> 2(Y+7) = 28

Verizon [17]

2y+14=28
-14=-14
—————-
2y=14
14/2=y
Y=7

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Circle O has a circumference of 36π cm. Circle O with radius r is shown. What is the length of the radius, r? 6 cm 18 cm 36 cm 7

Gwar [14]

Answer: 18 cm

Step-by-step explanation:

We know the circumference formula is C=2πr. Since our circumference is given in terms of π, we can easily figure out what the radius is.

36π=2πr [divide both sides by π to cancel out]

36=2r [divide both sides by 2]

r=18 cm

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How do you solve for 2y=x+3

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Your answer is
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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

Jason played golf at a rate of 6 holes in 24 minutes. How much longer would it take him to play 54 holes than 41 holes?

Mkey [24]

6 holes in 24 mins.
1 hole in 4 mins.
54 × 4= 216
41 × 4= 164
216-164= 52
So it would take 52 more minutes.

5 0

3 years ago