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3 years ago

Use the model to write the equation that represents the problem.

Mathematics

2 answers:

ella [17]3 years ago

8 0

Answer:

I wanna say 13 1/2 is the answer

Step-by-step explanation:

pishuonlain [190]3 years ago

6 0

Answer:

13 1/2

Step-by-step explanation:

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this is a blueprint drawing of the stage area at Millennium Park the length of one square on the grid is equal to 5 feet accordi

dmitriy555 [2]

Answer:

No, diagonal of square C is 70.71ft

Step-by-step explanation:

Given in the question that length of one square = 5 ft

Length of Square C = 10 small squares

1 square = 5 ft

10 square = 5ft * 10

= 50 ft

As we know that in a square all side are of equal length and make an angle of 90° at each corner.

So, by using pythagorus theorem

Diagonal of square² = side² + side²

Diagonal of square² = 2side²

Diagonal of square = √(2side²)

Diagonal of square = √2 (side)

where side = 50 ft

Diagonal of square = √2 (50)

Diagonal of square = 70.71 ft

6 0

2 years ago

Mr. Schmidt ordered 48 typewriters for his office. Each type writer cost $195. About how much did the type writers cost?

alexandr1967 [171]

One type writer = $195
48 Type writers = $195 x 48 = $9360

6 0

2 years ago

Read 2 more answers

Find the volume of the solid bounded by z = 2 - x2 - y2 and z = 1. Express your answer as a decimal rounded to the hundredths pl

butalik [34]

Answer:

The answer is " $(\frac{\pi}{2})$ ".

Step-by-step explanation:

$z = 2 - x^2 - y^2.........(1) \\\\z = 1.............(2)$

Let add equation 1 and 2:

Using formula: $x^2+y^2=1 \\\\$

convert to polar coordinates

$r=2$

$\theta \varepsilon (0=\pi)=z\\\\V=\int^{2\pi}_{\theta=0}\int^{1}_{\pi=0}\int^{z_{2}}_{z_1} r \ dr \d \theta\\\\$

$=\int^{2\pi}_{0}\int^{1}_{0} (Z_2-z_1)r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r) \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0} \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4}) \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\$