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3 years ago

Solve the inequality. 2(4 + 2x) < 5x + 5? X<_____

Mathematics

1 answer:

mixer [17]3 years ago

7 0

Answer:

x > 3

Step-by-step explanation:

Given

2(4 + 2x) < 5x + 5 ← distribute left side

8 + 4x < 5x + 5 ( subtract 5 from both sides )

3 + 4x < 5x ( subtract 4x from both sides )

3 < x ⇒ x > 3

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What is the value of x

Natali [406]

Answer:

x = 12

Step-by-step explanation:

The two angles are complementary since they form a 90 degree angle.

35+ 4x+7 = 90

Combine like terms

42 +4x = 90

Subtract 42 from each side

42-42 +4x =90-42

4x = 48

Divide each side by 4

4x/4 = 48/4

x = 12

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3 years ago

What is 4/3 subtracted by 2/5

yKpoI14uk [10]

In order to subtract fractions, we must find a common denominator, then subtract the numerators.

The common denominator (LCM(3, 5)) is 15.

We must multiply the first numerator by 5, because 3*5=15 and the second numerator by 3 because 5*3=15.

We get $\frac{20}{15} - \frac{6}{15}$ .

All we need to do now is subtract the numerators: $\frac{14}{15}$ .

The answer is 14/15.

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3 years ago

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A chef uses 2 tablespoons of olive oil for every 3 tablespoons of butter. How much olive oil is needed for each tablespoon of bu

BigorU [14]

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3 years ago

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Which expression is equivalent to (5^3)^2

Katarina [22]

Answer:

$(5^3)^2 = 5^6 = 15625$

<em>good luck, i hope this helps :)</em>

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3 years ago

At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

beks73 [17]

<h2>Answer:</h2>

$s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}$

and,

$v_{x}=1.39a_{x}+2.5$

<h2>Step-by-step explanation:</h2>

In the question,

Taking the elevation of pool along the y-axis, and length of the board along the x-axis.

On drawing the illustration in the co-ordinate system we get,

lₓ = 2 m

uₓ = 2.5 m/s

and,

$h_{y}=9.5\,m$

So,

From the equations of the laws of motion we can state that,

$s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}$

So,

On putting the values we can say that,

$s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=(0)t+\frac{1}{2}(9.8)t^{2}\\t^{2}=\frac{9.5}{4.9}\\t^{2}=1.93\\t=1.39\,s$

Now,

The <u>equation of the motion in the horizontal</u> can be given by,

$v_{x}=u_{x}+a_{x}t\\v_{x}=2.5+a_{x}(1.39)\\So,\\v_{x}=1.39a_{x}+2.5$

<em><u>Therefore, the equations of the motions in the horizontal and verticals are,</u></em>

$s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}\\9.5=4.9t^{2}$

and,

$v_{x}=1.39a_{x}+2.5$

6 0

2 years ago