All categories

3 years ago

Can someone help me i'm really lost

Mathematics

2 answers:

aniked [119]3 years ago

8 0

1,100m=850m+2000

First, subtract 850m from both sides.

1,100-850=250

250m=2000

Divide each side by 250

m=8

It will be 8 months when they paid the same amount.

1100(8)=8,800

Shaun paid 8,800 dollars

850(8)=6,800
6800+2000=8,800
Taliyah paid 8,800 dollars

8,800+8,800=17,600

They spent 17,600 together.

A) 8 months
B) 17,600 dollars

rusak2 [61]3 years ago

4 0

Its asking how much did they both spent, in total, in order to reach the same amount spent in part A

You might be interested in

consider a polynomial f(x)=ax^3 + bx^2 + x + 2/3.if x + 3 is a factor of f(x) and f(x) is divided by x + 2, then we get remainde

Elanso [62]

Answer:

a = 2/27

b = 13/27

Step-by-step explanation:

The given polynomial is presented as follows;

f(x) = a·x³ + b·x² + x + 2/3

Given that x + 3 is a factor, we have;

f(-3) = 0 = a·(-3)³ + b·(-3)² - 3 +2/3 = 0

-27·a + 9·b - 3 + 2/3 = 0

-27·a + 9·b = 7/3........(1)

Also we have

(a·x³ + b·x² + x + 2/3) ÷ (x + 2) the remainder = 5

Therefore;

a·(-2)³ + b·(-2)² + (-2) + 2/3 = 5

-8·a + 4·b - 2 + 2/3 = 5

-8·a + 4·b = 2 - 2/3 = 4/3........(2)

Multiplying equation (1) by 4/9 and subtracting it from equation (2), we have;

-8·a + 4·b - 4/9×(-27·a + 9·b) = 4/3 - 4/9 × 7/3

-8·a + 12·a = 8/27

4·a = 8/27

a = 2/27 ≈ 0.0741

imputing the a value in equation (1) gives;

-27×2/27 + 9·b = 7/3

-2 + 9·b = 7/3

9·b = 7/3 + 2 = 13/3

b = 13/27 ≈ 0.481.

6 0

3 years ago

Type the correct answer in each box. Spell all words correctly.

pogonyaev

Answer:

Charlie is incoerre

Step-by-step explanation:

Given that :

Line of sight to top of flagpole = 16 feets

Angle of elevation = 60°

The height of the flagpole :

Applying trigonometry to the drawing in the attached picture :

We use:

Sinα = opposite / hypotenus

This enables us to use the value of the hypotenus (line of sight) to obtain the missing height value (h)

Sin(60°) = h / 16

h = 16 * sin 60

h = 16 * 0.8660254

h = 13.856406

Hence flagpole is about 13.86 feets high

Charlie's reasoning is incorrect, the line of sight constitutes the hypotenus and not Adjacent and hence, shoul

3 0

3 years ago

Please help me answer this with the correct answer :)

BartSMP [9]

Pretty sure is 63 degrees since the one next to it is exactly the same and its 63 degrees

7 0

3 years ago

In the episode pertaining to the discovery of radium, all of the hypotheses turned out to be true.

Sunny_sXe [5.5K]

Answer:

In the episode pertaining to the discovery of radium, it is false that all of the hypotheses turned out to be true.

There are some hypotheses in the episode thay turned out to be false.

8 0

3 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago