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Nat2105 [25]
3 years ago
14

PLZZZZZZ HELP 61. Jill had the following scores on quizzes. 1 quiz: 1 3 quizzes: 2 4 quizzes: 3 4 quizzes: 4 1 quiz: 5 What is t

he mean score for Jill’s quizzes? Round to the nearest hundredth and show your work. Answer:
Mathematics
1 answer:
KIM [24]3 years ago
6 0
To find the mean add all the scores and divide your answer by the total amount of quizzes she took.  

Rounding to the nearest hundredth is the second number AFTER the decimal.  Look at the number to the right of the hundredths.  If it's five or higher, the number in the hundredths column increases by one.  If it's 4 or below, the number in the hundredths column stays the same
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A newspaper costs 80 pence. Tom buys a newspaper everyday. How much will Tom spend on newspapers for: four days? & seven?
Iteru [2.4K]

Answer:

four days: 320 pence

seven days: 560 pence

Step-by-step explanation:

Multiply the cost by the number of days.

80 pence× 4 days= 320 pence

80×4=320

80×7=560

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A line going through the points (30,45) and (60,30)​
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IrinaVladis [17]
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1/12+3/8y=5/12+5/8y , what does y equal
hram777 [196]
<span>1/12+3/8y=5/12+5/8y

</span><span>3/8y- 5/8 y = 5/12- 1/12
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4 0
3 years ago
Read 2 more answers
Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable
bekas [8.4K]

Answer:

\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

u=\frac{y}{x}

Multiply both sides by x:

ux=y

Second step: Differentiate the substitution:

u'x+u=y'

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}

u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of x^2 so I'm going to divide top and bottom by x^2:

u'x+u=\frac{u+u^2}{3+u^2}

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

u'x=\frac{u+u^2}{3+u^2}-u

Find a common denominator: Multiply second term on right hand side by \frac{3+u^2}{3+u^2}:

u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}

Combine fractions while also distributing u to terms in ( ):

u'x=\frac{u+u^2-3u-u^3}{3+u^2}

u'x=\frac{-u^3+u^2-2u}{3+u^2}

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

u' \frac{3+u^2}{-u^3+u^2-2u}x=1

Divide both sides by x:

\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}

Reorder the top a little of left hand side using the commutative property for addition:

\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0
3 years ago
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