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3 years ago

Each part of a race is 1/10 mile marker finished 5 parts of the race how far did Marshall race

Mathematics

2 answers:

exis [7]3 years ago

7 0

A half mile bc 1/10•5 = 5/10& thats 1/2

Nonamiya [84]3 years ago

3 0

5/10=1/2=0.5 these are the same answer but in different for the first is the original the second is the reduced form and the third is the decimal hope this helps hope i am brainliest i need one more

You might be interested in

Estimat the product of 32*36

7nadin3 [17]

32(40-4)
32*40 - 32*4
1,280 - 128=
1,152
therefor
32*36= 1,152

4 0

4 years ago

Some kids are selling lemonade for $1.50 per cup at a high

BaLLatris [955]

A function that would represent profit based on the number of cups of lemonade is Profit = 1.5n - 14

<u>Solution:</u>

Given, Some kids are selling lemonade for $1.50 per cup at a high school baseball game.

They spent $14 on all of the items needed for the lemonade stand (cups, lemonade, table oth, sign, etc)

We have to create a function that would represent profit based on the number of cups of lemonade

Now, let the number of cups sold be "n"

Then , we know that,<em> profit = selling price – cost price </em>

Profit = number of cups sold x price per cup – cost price

Profit = n x $ 1.5 – $ 14

Profit = 1.5n – 14

Hence, the function is Profit = 1.5n - 14

6 0

3 years ago

Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes

azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

The centre of the ellipse is at the origin and the X axis is the major axis

It passes through the points (-3, 1) and (2, -2)

The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

$\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

Given that the ellipse passes through the point (-3, 1)

Hence,

$\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1$

Cross multiplying we get,

9b² + a² = 1 ²× a²b²
a²b² = 9b² + a²

Multiply by 4 on both sides,

4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

$\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1$

Cross multiply,

4b² + 4a² = 1 × a²b²
a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

3a²b² = 32b²
3a² = 32
a² = 32/3----(3)

Substituting in 2,

32/3 × b² = 4b² + 4 × 32/3
32/3 b² = 4b² + 128/3
32/3 b² = (12b² + 128)/3
32b² = 12b² + 128
20b² = 128
b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

$\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1$

$\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1$

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>

8 0

3 years ago

1500 soldiers arrive at a training camp. A few soldiers desert the camp. The drill sergeants divide the remaining soldiers into

iren [92.7K]

Answer:

The number of deserters is 34.

Step-by-step explanation:

We have to calculate the number of desertors in a group of 1500 soldiers.

The sergeant divides in groups of different numbers and count the lefts over.

If he divide in groups of 5, he has on left over. The amount of soldiers grouped has to end in 5 or 0, so the total amount of soldiers has to end in 1 or 6.

If he divide in groups of 7, there are three left over. If we take 3, the number of soldiers gruoped in 7 has to end in 8 or 3. The only numbers bigger than 1400 that end in 8 or 3 and have 7 as common divider are 1428 and 1463.

If we add the 3 soldiers left over, we have 1431 and 1466 as the only possible amount of soldiers applying to the two conditions stated until now.

If he divide in groups of 11, there are three left over. We can test with the 2 numbers we stay:

$(1431-3)/11=1428/11=129.82\\\\(1466-3)/11=1463/11=133$