All categories

3 years ago

How to solve the formula for Vf in the formula of acceleration?

Mathematics

1 answer:

balandron [24]3 years ago

3 0

A = (Vf - Vi)/t
Multiply by t on both sides to clear the denominator
at = Vf - Vi
Add Vi to both sides.
at +Vi = Vf

You might be interested in

V square+6v-59=0<br> answer in one equation

Tema [17]

X= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
a=1
b=6
c=-59
x= $\frac{-6+/- \sqrt{(6)-4(1)(-59)} }{2(1)}$
x= $-3+ 2\sqrt{17}$ and $-3- 2\sqrt{17}$

8 0

3 years ago

Is the following statement true? Explain.

sukhopar [10]

Step-by-step explanation:

We can prove the statement is false by proof of contradiction:

We know that cos0° = 1 and cos90° = 0.

Let A = 0° and B = 90°.

Left-Hand Side:

cos(A + B) = cos(0° + 90°) = cos90° = 0.

Right-Hand Side:

cos(A) + cos(B) = cos(0°) + cos(90°)

= 1 + 0 = 1.

Since LHS =/= RHS, by proof of contradiction,

the statement is false.

4 0

3 years ago

Click this link to view O'NET's Skills section for Municipal Clerks Note that common skills are listed toward the top

Alika [10]

Writing, reading comprehension

3 0

3 years ago

Read 2 more answers

The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.

MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle = $3\times\ side$

$3\times\ side$ $=[tex]81\sqrt{3}$

$Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm$

Thus from the diagram , Side $AB=BC=AC= 27\sqrt{3} \ cm$

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus $\angle OBC = 30$ °

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

$tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2} } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\$

Thus ,apothem OE= 13.5 cm

Now for radius,

We consider ΔOBE

$cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB} \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm$