In half an hour josh ran 3/4 a mile. how many miles would Josh run in 1 hour?

Least common denominator (LCD) of 5/12 and 11/9

serious [3.7K]

Least common denominator is 36

5 0

3 years ago

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The work of a student to solve a set of equations is shown:

shtirl [24]

Option: step 2.

Because the student divided only the right side by 2. If you divide one side by a number you have to do the same with the other side to maintain the equality.

3 0

4 years ago

Answers this for me plzzz really quickly

Anna [14]

I think you have the right answer she paid $55 dollars for each chair

3 0

3 years ago

Read 2 more answers

A kite is designed on a rectangular grid with squares that measure 1cm by 1 cm. A hexagonal piece within the kite will be reserv

Nata [24]

Answer:

The answer is the first answer

P = 8 + 4√13 cm

A = 36 cm²

Step-by-step explanation:

* Lets study the figure

- Its a kite with two diagonals

- The shortest one is 12 cm

- The longest one is 26 ⇒ axis of symmetry of the kite

- the shortest diagonal divides the longest into two parts

- The smallest part is 8 cm and the largest one is 18 cm

* To find the area reserved for the logo divide

the hexagonal piece into two congruent trapezium

- The length of the two parallel bases are 4 cm and 8 cm and

its height is 3 cm

- The length of non-parallel bases can calculated by Pythagoras rule

∵ The lengths of the two perpendicular sides are 2 cm and 3 cm

- 3 cm is the height of the trapezium

- 2 cm its the difference between the 2 parallel bases ÷ 2

(8 - 4)/2 = 4/2 = 2 cm

∴ The length of the non-parallel base = √(2² + 3²) = √13

* Now we can find the area of the space reserved for the logo

- The area of the trapezium = (1/2)(b1 + b2) × h

∴ The area = (1/2)(4 + 8) × 3 = (1/2)(12)(3) = 18 cm²

∵ The space reserved for the logo are 2 trapezium

∴ The area reserved for the logo = 2 × 18 = 36 cm²

* The area of the reserved space for the logo = 36 cm²

* The perimeter of the reserved space for the logo is the

perimeter of the hexagon

∵ The lengths of the sides of the hexagon are:

4 cm , 4 cm , √13 cm , √13 cm , √13 cm , √13 cm

∴ The perimeter = 2(4) + 4(√13) = 8 + 4√13 cm

* The perimeter of the reserved space for the logo = 8 + 4√13 cm

4 0

3 years ago

Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant

Alex787 [66]

Find the intercepts for both planes.

Plane 1, x + y + 2z = 2:

$y=z=0\implies x=2\implies (2,0,0)$

$x=z=0\implies y=2\implies(0,2,0)$

$x=y=0\implies 2z=2\implies z=1\implies(0,0,1)$

Plane 2, 4x + 4y + z = 8:

$y=z=0\implies4x=8\implies x=2\implies(2,0,0)$

$x=z=0\implies4y=8\impliesy=2\implies(0,2,0)$

$x=y=0\implies z=8\implies(0,0,8)$

Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.

So the volume of the region is

$\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx$

$=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}$

3 0

3 years ago

In half an hour josh ran 3/4 a mile. how many miles would Josh run in 1 hour?​

In half an hour josh ran 3/4 a mile. how many miles would Josh run in 1 hour?