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2 years ago

An object has a mass of 400 g and a volume of 25 cm3. Find the density of the object in g/cm3.

Mathematics

1 answer:

Sunny_sXe [5.5K]2 years ago

4 0

Answer:

density is equal to mass / volume

Step-by-step explanation:

mass / volume

400/25

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Solve sint = cos2t, for –4π ≤ t ≤ –2π

e-lub [12.9K]

Answer:

-(23/6)π

-(19/6)π

-3π

Step-by-step explanation:

cos2t=cos^2t-sin^2t => cos^2t=1-sin^2t =>cos2t=1-2sin^2t

sint=1-2sin^2t, if x=sint, thenn we have 2x^2+x-1=0

Here x=0.5, x=-1, as x=sint => sint = -1 solutions for -4π to -2π is t=-3π,

sint = 0.5 solutions for -4π to -2π is t= -(23/6)*π, and t = -(19/6)*π

3 0

3 years ago

What is the amount of change? Show your work a. 85% of 210 b. 125% of 38

Stolb23 [73]

Basically you need to multiply the percent by the number.

7 0

3 years ago

Find the perimeter of pentagon STUVW WITH VERTICES S(0,0) T(3,-2) U(2,-5) V(-2,-5) W(-3,-2)

andrew11 [14]

Use the distance formula.

$\sqrt{( x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}$

Points S and W.
$\sqrt{(3)^2 + (2)^2}$

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points S and T
$\sqrt{(3 - 0)^2 + (-2 - 0)^2}$

$\sqrt{(3)^2 + (-2)^2}$

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points T and U
$\sqrt{(3 - 2)^2 + (-2 + 5)^2}$

$\sqrt{(1)^2 + (3)^2}$

$\sqrt{1+9}$

$\sqrt{10}$

~3.1

Points U and V
$\sqrt{(2+2)^2 + (-5 + 5)^2}$

$\sqrt{(4)^2 + (0)^2}$

$\sqrt{16}$

~4

Points V and W
$\sqrt{(-2+3)^2 + (-5 + 2)^2}$

$\sqrt{(1)^2 + (-3)^2}$

$\sqrt{2+9}$

$\sqrt{11}$

~3.3

Add all these together.

3.3 + 3.1 + 4 + 3.1 + 3.6
≈17

4 0

3 years ago

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for area of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side a = 10.

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$

So, radius of circle = $10-5\sqrt2$

8 0

2 years ago

Given the function f(x)=2xf(x)=2x, evaluate f(3)f(3).

kifflom [539]

Answer:

hope it helps you............

8 0

2 years ago