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3 years ago

Solve the equation tanxtan(x – 45°)=0 for 0° < x < 360°.

Mathematics

1 answer:

Lady_Fox [76]3 years ago

4 0

Answer:

\tan \left(\frac{\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}\right)

Step-by-step explanation:

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Write an equation for the line below. Show work

Gekata [30.6K]

x=3 is your answer. This is because it only crosses the x-axis and this occurs at positive 3. y=3 on the other hand would be a horizontal line that only touches positive 3 on the y-axis.

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2 years ago

How to find for 19 in the happy number

Gnoma [55]

Answer:

Just hope for the best

Step-by-step explanation:

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3 years ago

Read 2 more answers

3. Find the slope of the line<br> f(x)1/2x+8. Describe how the line slopes.

wariber [46]

Answer:

slope: $\frac{1}{2}$

Step-by-step explanation:

$f(x)=mx+b\\m=slope\\\\f(x)=\frac{1}{2} x+8\\\frac{(rise)}{(run)}=\frac{1}{2}$

the linear line has a positive slope that goes up by 1 unit and right by 2 units

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3 years ago

which has greater cenetic energy a car traveling 30.0 km/hr or one twice as heavy traveling at 15 km/hr?

iVinArrow [24]

Answer:

30 km/h car

Step-by-step explanation:

From analysis the car traveling at 30 km/h has greater kinetic energy

we can deduce it from the expression of kinetic energy which is

$KE=\frac{1}{2} mv^2$

Assuming the mass m= 1 kg

For the 30 km/h

$KE=\frac{1}{2}*1*30^2 \\\\KE=\frac{1}{2}*1*900\\\\\KE=450 J$

For the 15 km/h

$KE=\frac{1}{2}*2*15^2 \\\\ KE=\frac{1}{2}*2*225 \\\\\ KE=\frac{1}{2}*450 J\\\\\ KE=225 J$

Though the kinetic energy is a function of mass and velocity, but from our analysis the faster moving object has more KE

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3 years ago

Point D is at (3,-4) and is translated 5 units up

adoni [48]

Answer:

(-1,1)

Step-by-step explanation:

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3 years ago