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adell [148]
2 years ago
5

Write as an expression " 6 times smaller than 42"​

Mathematics
1 answer:
Mademuasel [1]2 years ago
5 0

Answer:

42/6 or 7

Step-by-step explanation:

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Can anyone help me out here? 30 POINTS for anyone that can give me the answer and explain how they got it...
charle [14.2K]

Answer:


Step-by-step explanation:

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5 0
2 years ago
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for competition,the cheerleading team,is preparing a dance routine that must last 4 minutes,with a variation of +- 5 seconds. fi
Sonbull [250]

Ideal timeline of the dance routine = 4 minutes = 4 × 60 seconds = 240 seconds

Variation allowed in the dance routine timeline = +- 5 seconds

Let the timeline of the dance routine be T

⇒ 240 seconds - 5 seconds < T < 240 seconds + 5 seconds

⇒ 235 seconds < T < 245 seconds

⇒ \frac{235}{60} minutes < T < \frac{245}{60} minutes

⇒ 3.92 minutes < T < 4.08 minutes

So the least possible time of the dance routine can be 3.92 minutes (or 235 seconds) and the greatest possible time of the dance routine can be 4.08 minutes (or 245 seconds)

3 0
3 years ago
Can anyone help me with number 3 please
Gnoma [55]
Answer:


I think is (B)
4 0
2 years ago
Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,
Dahasolnce [82]

Answer:

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = \frac{a . i}{|a| . |i|}               ---------------------(i)

cos β = \frac{a.j}{|a||j|}               ---------------------(ii)

cos γ = \frac{a.k}{|a|.|k|}             ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i <em>is just the x component of the vector</em>]

a . j = 1            [<em>the y component of the vector</em>]

a . k = 4          [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = \sqrt{5^2 + 1^2 + 4^2}

|a| = \sqrt{25 + 1 + 16}

|a| = \sqrt{42}

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = \frac{5}{\sqrt{42} }

cos β =  \frac{1}{\sqrt{42} }              

cos γ =  \frac{4}{\sqrt{42} }

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

α =  cos⁻¹ ( \frac{5}{\sqrt{42} } )

α =  cos⁻¹ (\frac{5}{6.481} )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

β = cos⁻¹ ( \frac{1}{\sqrt{42} } )

β = cos⁻¹ ( \frac{1}{6.481 } )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

γ = cos⁻¹ (\frac{4}{\sqrt{42} })

γ = cos⁻¹ (\frac{4}{6.481})

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

<u>Conclusion:</u>

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

3 0
3 years ago
If a certain cannon is fired from a height of 8.8 meters above the​ ground, at a certain​ angle, the height of the cannonball ab
Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

h(t)=-4.9t^2+30.5t+8.8

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

-4.9t^2+30.5t+8.8=0

Multiply both sides by 10.

-49t^2+305t+88=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are: x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Substitute a = -49, b = 305 and c=88

t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}

Ignore the negative value of t as time can't be a negative number.

Thus,

t=\frac{305+\sqrt{110273}}{98}\approx6.50

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

6 0
3 years ago
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