Ask question

Ask question

All categories

3 years ago

12

Yolanda participated In a walkathon in which each kilometer walked raised $10 for charity, her goal was to raise more more than

$300 on Saturday and Sunday she raised $50 on Saturday which inequality is it?

2 answers:

Daniel [21]3 years ago

8 0

X= amount of money needed on sunday

50+x(greater or equal to)300

labwork [276]3 years ago

7 0

Answer: The required inequality will be 50+x>$300

Step-by-step explanation:

Since we have given that

Amount raised for charity = $10

Amount raised on Saturday = $50

Amount raised on both Saturday and Sunday is more than $300.

Let the amount raised on Sunday be 'x'.

So, Inequality becomes,

Saturday + Sunday > $300

50+x>$300

x>$300-$50

x>$250

So, On Sunday he must raised more than $250.

Hence, the required inequality will be

50+x>$300

You might be interested in

I need help now btw i am in middle school

vodomira [7]

Answer: x= 3

Step-by-step explanation:

2x + 5 = 11

2x = 6

x = 3

5 0

3 years ago

Read 2 more answers

11. If f(x)=4x-9, what is the equation for f^-1(x)?

FrozenT [24]

F(x) = 4x - 9

let f(x) = y, this implies that x = f⁻¹(y)

y = 4x - 9 Let us solve for x.

4x - 9 = y

4x = y + 9

x = (y + 9)/4

Recall that x = f⁻¹(y),

x = (y + 9)/4

f⁻¹(y) = (y + 9)/4

That means that for f⁻¹(x)

f⁻¹(x) = (x + 9)/4

Hope this explains it.

3 0

3 years ago

Explain the term <br><img src="https://tex.z-dn.net/?f=%20%5Cmathbb%20%7B%5Cgreen%7Babolute%20%5C%3A%20%7B%20%5Cpink%7Bmeasure%7

Setler [38]

Answer:

<h2> abolute measure </h2>

it is quantity measured and same unit data and also as an original data it's is an original data expressed and unit of series

<h2>range </h2>

largest value difference to smallest value in data

<h2>Quartiles </h2>

it is an semi and half of range

<h2>mean </h2>

also as arithmetic mean as more deviation of range

3 0

2 years ago

The coordinates of the vertices of a rectangle are (−3, 4) , (7, 2) , (6, −3) , and (−4, −1) . What is the perimeter of the rect

ki77a [65]

Answrer

Find out the what is the perimeter of the rectangle .

To prove

Now as shown in the figure.

Name the coordinates as.

A(−3, 4) ,B (7, 2) , C(6, −3) , and D(−4, −1) .

In rectangle opposite sides are equal.

Thus

AB = DC

AD = BC

Formula

$Disatnce\ formula = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2}}$

Now the points A(−3, 4) and B(7, 2)

$AB = \sqrt{(7- (-3))^{2} +(2- 4)^{2}}$

$AB = \sqrt{(10)^{2} +(-2)^{2}}$

$AB = \sqrt{100+4}$

$AB = \sqrt{104}$

$AB = 2\sqrt{26}\units$

Thus

$CD= 2\sqrt{26}\units$

Now the points

A (−3, 4) , D (−4, −1)

$AD = \sqrt{(-4 - (-3))^{2} +(-1- 4)^{2}}$

$AD = \sqrt{(-1)^{2} +(-5)^{2}}$

$AD = \sqrt{1 + 25}$

$AD = \sqrt{26}\units$

Thus

$BC = \sqrt{26}\units$

Formula

Perimeter of rectangle = 2 (Length + Breadth)

Here

$Length = 2\sqrt{26}\ units$

$Breadth = \sqrt{26}\ units$

$Perimeter\ of\ rectangle = 2(2\sqrt{26} +\sqrt{26})$

$Perimeter\ of\ rectangle = 2(3\sqrt{26})$

$Perimeter\ of\ rectangle = 6\sqrt{26}$

$\sqrt{26} = 5.1 (Approx)$

$Perimeter\ of\ rectangle = 6\times 5.1$

Perimeter of a rectangle = 30.6 units.

Therefore the perimeter of a rectangle is 30.6 units.

8 0

3 years ago

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 0

2 years ago