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MrMuchimi
3 years ago
9

1 quart of water 1 pint of juice and 9 ounces of honey equals how much total pints, cups, and ounces

Mathematics
1 answer:
Georgia [21]3 years ago
5 0
57 ounces because every quart has 2 pints and every pint has 16 ounces
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Write an addition equation and multiplication equation that each have a solution of -5.
Nookie1986 [14]
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3 years ago
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Prove that: 5^31–5^29 is divisible by 100.
ANEK [815]

Answer:

See explanation

Step-by-step explanation:

Consider the expression

5^{31}-5^{29}

First, factor it:

5^{31}-5^{29}=5^{29}\cdot(5^2-1)=5^{29}\cdot (25-1)=24\cdot 5^{29}

Note that

100=25\cdot 4

Then

5^{31}-5^{29}=24\cdot 5^{29}=6\cdot 4\cdot 25\cdot 5^{27}=6\cdot 100\cdot 5^{27}

This shows that number 100 is a factor of the expression 5^{31}-5^{29} and, therefore, this expression is divisible by 100.

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3 years ago
What is the circumference of the circle defined by x^2+6x +y^2-12y-4=0.
KengaRu [80]

Answer:

44.0

Step-by-step explanation:

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5 0
3 years ago
Bayside Insurance offers two health plans. Under plan A, Giselle
LiRa [457]

Answer:

Step-by-step explanation:

Let x represent the amount of medical bills that Giselle has to pay.

Assume she has over $160 in bills,

it means that

x > 160

Under plan A, Giselle would have to pay the first $110 of her medical bills, plus 35% of the rest. Therefore, she will have to pay for

35/100 × (x - 110) = 0.35(x - 110)

Under plan B, Giselle would pay the first $160, but only 20% of the rest. Therefore, she will have to pay for

20/100 × (x - 160) = 0.2(x - 160)

Therefore, the amount of medical bills that plan B will save is

0.35(x - 110) - 0.2(x - 160) = 0.35x - 38.5 - 0.2x + 32

0.15x + 6.5

Substituting x = 160 into 0.15x + 6.5, it becomes

0.15 × 160 + 6.5 = $30.5

Total bills would be 160 + 30.5 = 190.5

Therefore, Giselle would save $30.5 with plan B if she had more than $190.5 in bills.

4 0
3 years ago
What is the difference between bernoulii equation and riccati equation
melisa1 [442]
The Bernoulli equation is almost identical to the standard linear ODE.

y'=P(x)y+Q(x)y^n

Compare to the basic linear ODE,

y'=P(x)y+Q(x)

Meanwhile, the Riccati equation takes the form

y'=P(x)+Q(x)y+R(x)y^2

which in special cases is of Bernoulli type if P(x)=0, and linear if R(x)=0. But in general each type takes a different method to solve. From now on, I'll abbreviate the coefficient functions as P,Q,R for brevity.

For Bernoulli equations, the standard approach is to write

y'=Py+Qy^n
y^{-n}y'=Py^{1-n}+Q

and substitute v=y^{1-n}. This makes v'=(1-n)y^{-n}y', so the ODE is rewritten as

\dfrac1{1-n}v'=Pv+Q

and the equation is now linear in v.

The Riccati equation, on the other hand, requires a different substitution. Set v=Ry, so that v'=R'y+Ry'=R'\dfrac vR+Ry'. Then you have

y'=P+Qy+Ry^2
\dfrac{v'-R'\dfrac vR}R=P+Q\dfrac vR+R\dfrac{v^2}{R^2}
v'=PR+\left(Q+\dfrac{R'}R\right)v+v^2

Next, setting v=\dfrac{u'}u, so that v'=\dfrac{uu''-(u')^2}{u^2}, allows you to write this as a linear second-order equation. You have

\dfrac{uu''-(u')^2}{u^2}=PR+\left(Q+\dfrac{R'}R\right)\dfrac{u'}u+\dfrac{(u')^2}{u^2}
u''-\left(Q+\dfrac{R'}R\right)u'+PRu=0
u''+Su'+Tu=0

where S=-\left(Q+\dfrac{R'}R\right) and T=PR.
3 0
3 years ago
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