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2 years ago

X 2x+24 Equation Plz help

Mathematics

2 answers:

Nataly_w [17]2 years ago

8 0

Answer:

$x + 2x + 24 = 180 \\ 3x = 180 - 24 \\ 3x = 156 \\ x = 52$

kvasek [131]2 years ago

4 0

Answer: x = 52° and 2x + 24 = 128°

Steps:
180 = x + (2x + 24)
180 = x + 2x + 24
180 = 3x + 24
180 - 24 = 3x
156 = 3x
52 = x

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As all of the bundles have the same content inside, so assuming that there is x number of Fiction books and y number of Non-fiction books in each bundle.

Let n be the total number of bundles that my team can send.

There are 64 Fiction books, so

nx=64 ...(i)

Or x=64/n ...(ii)

also, there are 48 Non-Fiction books, so

ny=48 ...(iii)

Or y=48/n ...(iv)

Observing that the numbers x, y, and n are counting numbers and from equations (i) and (iii), n is the common factor of 64 and 48.

The possible common factors of 64 and 48 are,

n=1,2,4,8, and 16.

So, my team can send 1,2,3,4,8 or 16 bundles of books.

Now, from equations (ii) and (iv),

For n=1:

x=64/1=64

y=48/1=48

So, for 1 bundle the number of Fiction and Non-fictions books are 64 and 48 respectively.

For n=2:

x=64/2=32

y=48/2=48

So, for 2 bundles, the number of Fiction and Non-fictions books are 32 and 24 respectively.

For n=4:

x=64/4=16

y=48/4=12

So, for 4 bundles, the number of Fiction and Non-fictions books are 16 and 12 respectively.

For n=8:

x=64/8=8

y=48/8=6

So, for 8 bundles, the number of Fiction and Non-fictions books are 8 and 6 respectively.

For n=16:

x=64/16=4

y=48/16=3

So, for 16 bundles, the number of Fiction and Non-fictions books are 4 and 3 respectively.

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3 years ago

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Answer:

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tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

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And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

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$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

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Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

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