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2 years ago

How can properties of operations be used to solve problems involving integers and rational numbers

1 answer:

7 0

Answer: The properties of operations can be used to solve problems involving integers because it shows different way to find the answer.

Step-by-step explanation:

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6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the

MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

$a_1$ the horizontal distance between the first observer and the ballonist

$a_2$ the horizontal distance between the second observer and the ballonist

$\alpha _1$ and $\alpha _2$ the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

$S=a*h$ (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

$S=S_1+S_2$ (equation 2)

$S_1=a_1*h$

But we can write $S_1$ in terms of $\alpha _1$ , like this:

$\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}$

And for $S_2$ will be the same:

$S_2=\frac{h^{2} }{\tan(\alpha _2)}$

Replacing in the equation 2:

$S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})$

And replacing in the equation 1:

$h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}$

So, we can replace all the known data in the last equation:

$h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft$

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

6 0

2 years ago

Quadrilateral QRST is similar to quadrilateral EDGF. What's the scale factor from EDGF to QRST

mart [117]

Answer:

C. $\frac{6}{5}$

Step-by-step explanation:

The ratio of the corresponding sides of a polygon is said to be equal. The ratio gives us the scale factor.

Therefore, assuming that quadrilateral QRST was the original polygon that was scaled up to give a similar polygon, EDGF, the scale factor would be expressed as $\frac{42}{35}$

The scale factor = $\frac{42}{35} = \frac{6*7}{5*7} = \frac{6}{5}$

5 0

2 years ago

Ashley plants a tree 45cm tall it grows by 10 cm each year. Find the amount ashley gets in the 5th year?

Genrish500 [490]

Answer:

95cm

Step-by-step explanation:

45cm is the initial height

each year grow 10cm

growing time 5 years

45+(10 × 5)=95 cm

4 0

2 years ago

For the school's sports day, a group of students prepared 12 1/2 litres of lemonade. At the end of the day they had 2 5/8 litres

Hoochie [10]

Given :

For the school's sports day, a group of students prepared 12 1/2 litres of lemonade. At the end of the day they had 2 5/8 litres left over.

To Find :

How many litres of lemonade were sold.

Solution :

Initial amount of lemonade, I = 12 1/2 = 25/2 litres.

Final amount of lemonade, F = 2 5/8 = 21/8 litres.

Amount of lemonade sold, A = I - F

A = 25/2 - 21/8 litres

A = 9.875 litres

Therefore, 9.875 litres of lemonade were sold.

Hence, this is the required solution.

7 0

2 years ago

A contractor has submitted bids on three state jobs: an office building, a theater, and a parking garage. State rules do not all

german

Answer:

The following are the answer to this question:

Step-by-step explanation:

In the given question the numeric value is missing which is defined in the attached file please fine it.

Calculating the probability of the distribution for x:

$\to f(x) = 0.19\ for \ x=14\\\\\to f(x) = 0.29 \ for\ x=7\\\\\to f(x) = 0.38\ for \ x=1\\\\\to f(x)=0.14 \ for \ x=0\\$

The formula for calculating the mean value:

$\bold{ E(X)= x \times f(x)}$

$=14 \times 0.19+7 \times 0.29+1 \times 0.38+0\times 0.14\\\\=2.66 + 2.03+0.38+ 0\\\\=5.07$

$\bold{E(X^2) = x^2 \times f(x)}$

$=14^2 \times 0.19+7^2 \times 0.29+1^2 \times 0.38+0^2 \times 0.14 \\\\=196 \times 0.19+ 49 \times 0.29+1 \times 0.38+0 \times 0.14\\\\= 37.24+ 14.21+ 0.38+0 \\\\=51.83$