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Alborosie
3 years ago
9

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
what do I do on C it says assess the reasonableness of your answer in B and use your estimate from A to explain
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I think that the answer is B<span />
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3 years ago
Your famous cookie recipe calls for
damaskus [11]

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I GOT -42 SO LIKE 4 1/2

Step-by-step explanation:

3 0
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
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