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fredd [130]
3 years ago
6

Would you rather become successful in school or become successful in life?

Mathematics
2 answers:
FrozenT [24]3 years ago
5 0
Life. Harry Styles dropped out of school and look where he is now
Illusion [34]3 years ago
5 0

Answer:

Successful in life

Step-by-step explanation:

The only reason I'm trying to be successful in school is so that i can be successful in life, but if i had a choice of being successful in life without school, I'll take it

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it takes 0.75 cups of sugar to make a batch of lemon cookies i have 5.5 cups of sugar how mant batches can i make *answer has to
Feliz [49]

Answer:

i

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
the half-time life of strontium-85 is 64.84 days. how much of a 40 mg sample would be left after 180 days​
Fynjy0 [20]

Answer:

  about 5.84 mg

Step-by-step explanation:

The amount remaining after t days can be written as ...

  remaining = initialAmount · (1/2)^(t/(halfLife))

Filling in the given numbers, you have

  remaining = (40 mg)·(1/2)^(180/64.84) ≈ 5.84 mg

About 5.84 mg would be left after 180 days.

7 0
3 years ago
Leave the answer as an improper fraction​
Ghella [55]

Hey there!

ANSWER: \frac{23}{20}

EXPLANATION:

To find the answer to your question, you will need to add.

\frac{3}{4}+\frac{2}{5}

First, multiply the denominator.

5*4=20

So the denominator will be 20.

Now move on to the numerator. Multiply 5 by 3.

5*3=15

Now multiply 4 bu 2.

4*2=8

If you want to get the numerator, add what you got after multiply.

15+8=23

The numerator is 23. So now let's complete the fraction.

\frac{23}{20}

This is your answer!

Hope this helps!

-TestedHyperr

3 0
3 years ago
A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so tha
Arlecino [84]

Answer:

A) F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }with a = \frac{2\Delta x}{t^2}

B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box

Step-by-step explanation:

A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).

In the y-axis, the box does not move, therefore:

\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha) (1)

where \alpha is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration a that can be calculated as a function of the given displacement and time interval as:

a = \frac{2\Delta x}{t^2} and the force equation is:

\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma

now substituting N from equation (1):

F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}

and putting the expression for the acceleration:

\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}} (2)

which is the requested expression

B) Substituting the values in (2) and using g=9.8\ \rm{ms^{-2}}:

F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}

The gravitational force acting on the box is:

F_g=mg=50\cdot9.8=490\ \rm{N}

F/F_g = (117.634/490)\cdot 100 = 24 \%

Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.

8 0
4 years ago
The figure on the left represents a scale drawing of the figure on the right. What is the scale?
docker41 [41]

Answer:

\frac{1}{90}

Step-by-step explanation:

Before calculating the scale we require the dimensions to be in the same units.

Using the conversion

1 yard = 3 ft and

1 foot = 12 inches, then

5 yards = 5 × 3 × 12 = 180 inches

The scale is then

2 in : 180 in ← divide both quantities by 2

= 1 : 90

= \frac{1}{90}

4 0
4 years ago
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