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lana66690 [7]
2 years ago
11

Where are mc donalds

Mathematics
2 answers:
Blizzard [7]2 years ago
7 0

Answer:

Not in Russia

Step-by-step explanation:

McDonald's left Russia following the violations of human rights

cluponka [151]2 years ago
5 0

Answer:

all over the world

Step-by-step explanation:

use g00g!e maps

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What is the sum?
Nezavi [6.7K]

Answer:

Its B on e2020

Step-by-step explanation:

correct on quiz

4 0
3 years ago
Read 2 more answers
Chelsea bought a computer on Monday for $1,300. Its value is predicted to decrease by $250 per year. Her brother George also bou
Elenna [48]

Answer:

The correct answer is B. George’s computer is expected to have a value of $25 greater.

Step-by-step explanation:

Since Chelsea bought a computer on Monday for $ 1,300, and its value is predicted to decrease by $ 250 per year, while her brother George also bought a computer on Monday, and the function g (x) = 1,100 - 175x predicts how the value of his computer is expected to change after x years, to determine whose computer is expected to have a greater value when it is 3 years old, and how much greater will it be, the following calculation must be performed:

Chelsea:

1,300 - (250 x 3) = X

1,300 - 750 = X

550 = X

George:

1,100 - (175 x 3) = X

1,100 - 525 = X

575 = X

Therefore, George’s computer is expected to have a value of $ 25 greater.

7 0
3 years ago
Max spent $53 and now has no money left. He had $ before his purchase.
Yuri [45]
Can you restate it or rewrite it? It doesn't have enough for me to answer this.
8 0
3 years ago
An ice cream store has the pricing shown below. You want to determine the best value. The height of the cone is 4.5 in and the d
nignag [31]

Solution:

Step 1:

We will calculate the volume of ice cream in the single scoop

The volume of the ice cream will be

\begin{gathered} V=\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3 \\ r=\frac{2in}{2}=1in(cone) \\ h=4.5in \\ r=\frac{3in}{2}=1.5in(radius\text{ of the hemisphere\rparen} \end{gathered}

By substituting the values, we will have

\begin{gathered} V=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^{3} \\ V=\frac{1}{3}\times\frac{22}{7}\times1^2\times4.5+\frac{2}{3}\times\frac{22}{7}\times1.5^3 \\ V=\frac{33}{7}+\frac{99}{14} \\ V=\frac{165}{14} \\ V=11.79in^3 \end{gathered}

Step 2:

We will use the formula below to calculate the volume of the two scoops of ic cream

\begin{gathered} V=\frac{1}{3}\pi r^2h+\frac{4}{3}\pi r^3 \\ V=\frac{1}{3}\times\frac{22}{7}\times1^2\times4.5in+\frac{4}{3}\times\frac{22}{7}\times1.5^3 \\ V=\frac{33}{7}+\frac{99}{7} \\ V=\frac{132}{7} \\ V=18.86in^3 \end{gathered}

Step 3:

We will use the formula below to calculate the volume of the three scoops of ic cream

\begin{gathered} V=\frac{1}{3}\pi r^2h+\frac{6}{3}\pi r^3 \\ V=\frac{1}{3}\times\frac{22}{7}\times1^2\times4.5+2\times\frac{22}{7}\times1.5^3 \\ V=\frac{33}{7}+\frac{297}{14} \\ V=\frac{363}{14} \\ V=25.93in^3 \end{gathered}

For the first ice cream with one scoop

\begin{gathered} 1in^3=\frac{3.50}{11.79} \\ 1in^3=\text{ \$}0.30 \end{gathered}

For the second ice cream with two scoops

\begin{gathered} 1in^3=\frac{4.50}{18.86} \\ 1in^3=\text{ \$}0.24 \end{gathered}

For the third ice cream with three scoops

\begin{gathered} 1in^3=\frac{5.50}{25.93} \\ 1in^3=\text{ \$}0.21 \end{gathered}

Hence,

The final answer is

The triple sold at $5.50 has the best value because it has the lowest price of $0.21 per cubic inch of the ice cream

3 0
1 year ago
A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
3 years ago
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