<img src="https://tex.z-dn.net/?f=%20%7B3%7D%5E%7B4%7D%20%20%5Ctimes%20%20%7B3%7D%5E%7B5%20%7D%20%20%2B2%285%20-%202%20%5Ctimes%

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2 years ago

205%29%20-%20%20%20%7B%284%20%2B%202%29%7D%5E%7B2%7D%20" id="TexFormula1" title=" {3}^{4} \times {3}^{5 } +2(5 - 2 \times 5) - {(4 + 2)}^{2} " alt=" {3}^{4} \times {3}^{5 } +2(5 - 2 \times 5) - {(4 + 2)}^{2} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

neonofarm [45]2 years ago

7 0

$3 {}^{4} \times {3}^{5} + 2(5 - 2 \times 5) - (4 + 2) \\ = 3 {}^{4 + 5} + 2(5 - 10) - (6) \\ = 3 {}^{9} + 2( - 5) - 6 \\ = 3 {}^{9} - 10 - 6 \\ = 19667$

3^4 × 3^5 Is the same as 3^9.

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Mr. Kennedy raise beef cattle the stock tank near the windmill has a diameter of 3 m what it is for the water is 1.5 m deep how

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Answer:

10.6

Step-by-step explanation:

radius = 1.5 height = 1.5 so it's 10.6

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3 years ago

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

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)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

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

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

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

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

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3 years ago

Is 18-35 an expression or an equation

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Answer:

It is an expression

Step-by-step explanation:

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Hope this helps!! :)

6 0

3 years ago

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Vinil7 [7]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

PLEASE HELP ME IM STRUGGLING!!!

Kitty [74]

Answer:

The required answer is $c=7\sqrt{3}$

Therefore the number in green box should be 7.

Step-by-step explanation:

Given:

AB = 7√2

AD = a , BD = b , DC = c , AC = d

∠B = 45°, ∠C = 30°

To Find:

c = ?

Solution:

In Right Angle Triangle ABD Sine identity we have

$\sin B = \dfrac{\textrm{side opposite to angle B}}{Hypotenuse}\\$

Substituting the values we get

$\sin 45 = \dfrac{AD}{AB}= \dfrac{a}{7\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}= \dfrac{a}{7\sqrt{2}}\\\\\therefore a=7$

Now in Triangle ADC Tangent identity we have

$\tan C = \dfrac{\textrm{side opposite to angle C}}{\textrm{side adjacent to angle C}}$

Substituting the values we get

$\tan 30 = \dfrac{AD}{DC}= \dfrac{a}{c}\\\\\dfrac{1}{\sqrt{3}}=\dfrac{7}{c}\\\\\therefore c=7\sqrt{3}$

The required answer is $c=7\sqrt{3}$

8 0

3 years ago