All categories

3 years ago

Here is my question:

Mathematics

1 answer:

sergiy2304 [10]3 years ago

6 0

Answer:

$\frac{2}{3} - \frac{1}{2} = \frac{1}{6} \times \frac{8}{9} = \frac{4}{27}$

You might be interested in

<img src="https://tex.z-dn.net/?f=5%20%2B%20%20%5Cfrac%7Bx%20-%202%7D%7B3%7D%20%20%3D%20%20%5Cfrac%7Bx%20%2B%203%7D%7B8%7D%20" i

sammy [17]

First multiply each term by 24(the common denominator found by 8×3) to remove the fractions and make things easier.
$24(5) + 24( \frac{x - 2}{3} ) = 24( \frac{x + 3}{8} )$
This will give you,
$120 + 8(x - 2) = 3(x + 3)$
Then just continue to simplify and isolate the variable.
$120 + 8x - 16 = 3x + 9$
$8x + 104 = 3x + 9$
$5x = - 95$
$x = - 19$

7 0

3 years ago

g If a snowball melts so that its surface area decreases at a rate of 4 cm2/min, find the rate (in cm/min) at which the diameter

maks197457 [2]

Answer:

Rate of change of the diameter is equal to 0.053cm/min

Step-by-step explanation:

A snowball is of the shape of a sphere

Surface area of Sphere (S)= $4*pi*R^2$

Rate of change of surface area

dS/dt = -4 $cm^2/min$

the negative sign indicates the surface area is decreasing.

Radius = d/2 where d represents diameter

using this is the surface area equation

S = $4*pi*(d/2)^2$

= $pi*d^2$

dS/dt = 2*pi*d*d(d)/dt

4 = 2*pi*d*(d)/dt

d(d)/dt = 2/12*pi= $\frac{1}{6pi}$ cm/min

so the rate of change of the diameter is equal to 0.053cm/min

5 0

3 years ago

which weighs more a watermelon that weighs 7.5 kilograms or a baby that weighs 12 pounds? note that 1 pound is about 0.45 kilogr

Sonbull [250]

Answer:

Da watermelon

5 0

3 years ago

Which function has a minimum and is transformed to the right and down from the parent function, f(x) = x2?

Aliun [14]

Answer:

The last choice is the one you want

Step-by-step explanation:

First of all, a parabola has a minimum value if it is a positive parabola, one that opens upward. The first and the third parabolas are negative so they open upside down. That leaves us with choices 2 and 4. We find the side to side and up or down movement by finishing the completion of the square that has already been started for us. Do this by factoring what's inside the parenthesis into a perfect square binomial.

The second one factored becomes:

$g(x)=4(x-3)^2+1$

which reflects a shift to the right 3 and up 1. Not what we are asked to find.

The fourth one factored becomes:

$g(x)=8(x-3)^2-5$

which reflects a shift to the right 3 and down 5. That's what we want!

8 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

3 years ago