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3 years ago

Barbara gets 6 pizza to divide equally among 4 people. How much of a pizza can each person ?

Mathematics

2 answers:

lana [24]3 years ago

8 0

Ok, well, had to edit this because i was being blond. SO yeah ignore this. Sorry :(

kozerog [31]3 years ago

3 0

Your key word is divide. You'll want to divide your 6 pizzas by your four friends, than most likely convert it to a fraction (unless your teacher accepts decimals.), message if you need more help.

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Uv = 8 and wx=5 find the missing measure of a rhombus

Harlamova29_29 [7]

Answer:

TU = UV = VW = WT = 8
WX = UX = 5
XV = XT = √39

Step-by-step explanation:

At first we should know the following about the rhombus

1) All sides are congruent

2) The diagonals are perpendicular and bisects each other.

Given UV = 8 and WX = 5

So, according to (1)

TU = UV = VW = WT = 8

And according to (2)

WX = UX = 5

And ΔWXV is a right triangle at ∠x

So, XV² = WV² - WX² = 8² - 5² = 64 - 25 = 39

∴XV = √39

So, XV = XT = √39

So, the missing measure of a rhombus are as following:

TU = UV = VW = WT = 8
WX = UX = 5
XV = XT = √39

3 0

3 years ago

Wyatt is enlarging a rectangular drawing using a scale factor of 2.5.<br> 7.4 in<br> 5 in

omeli [17]

It is 2.4 x 2.5 which equals 6 the answer is 6

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3 years ago

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poizon [28]

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3 years ago

Your famous cookie recipe calls for

damaskus [11]

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3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago