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Ilya [14]
3 years ago
10

How many solutions does -2z+2+4z=16/5+2z-6/5

Mathematics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

Infinite. Please Mark Brainliest, this took a lot of effort.

Step-by-step explanation:

After going through the process of solving, you get 0=0, meaning true for all z

-2z+2+4z=16/5+2z-6/5

Group Like Terms

-2z+4z+2=16/5+2z-6/5

Add similar terms

2z+2=16/5+2z-6/5

Group Like Terms

2z+2=2z-6/5+16/5

<em>Combine the Fractions</em>

Apply the rule a/c plus or minus b over c equals a plus or minus b over c

=(-6+16)/5

Add/Subtract the numbers

=10/5

Divide the numbers

=2

2z+2=2z+2

Subtract 2 from each side

2z+2-2=2z+2-2

Simplify

2z=2z

Subtract 2z from each side

2z-2z=2z-2z

Simplify

0=0

Both sides are equal, true for all z

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The quantity demanded x for a certain brand of MP3 players is 100 units when the unit price p is set at $80. The quantity demand
Cerrena [4.2K]

Answer:p = (2100 - x)/25

Step-by-step explanation:

According to the law of demand, when the price,p of the brand of MP3 players is is high, the quantity,x of the brand of MP3 players demanded would be low and when the price,p of the brand of MP3 players is is low, the quantity,x of the brand of MP3 players demanded would be high.

To derive the demand equation, we would apply the slope intercept form of equation which is expressed as

y = mx + c

Where

m = slope

c = intercept

The slope, m would be

(y2 - y1)/(x2 - x1)

Slope = (1100 - 100)/(40 - 80) = 1000/-40

Slope = - 25

To find the y intercept, we would substitute m = - 25, y = 1100 and x = 40 into y = mx + c. It becomes

1100 = - 25 × 40 + c

1100 = - 1000 + c

c = 1100 + 1000 = 2100

y = - 25x + 2100

Therefore, the demand equation is

x = - 25p + 2100

25p = 2100 - x

p = (2100 - x)/25

4 0
2 years ago
Which situation is represented by the equation x+10=26?
zhuklara [117]

Answer:

16

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
For the following telescoping series, find a formula for the nth term of the sequence of partial sums
gtnhenbr [62]

I'm guessing the sum is supposed to be

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}

Split the summand into partial fractions:

\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}

1=a(5k+4)+b(5k-1)

If k=-\frac45, then

1=b(-4-1)\implies b=-\frac15

If k=\frac15, then

1=a(1+4)\implies a=\frac15

This means

\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}

Consider the nth partial sum of the series:

S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)

The sum telescopes so that

S_n=\dfrac2{14}-\dfrac2{5n+4}

and as n\to\infty, the second term vanishes and leaves us with

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17

7 0
3 years ago
Solve E = mc2 for c. A) c = E m B) c = m E C) c = + m E D) c = + E m
S_A_V [24]

C=\sqrt{E/M}  

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6 0
3 years ago
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kumpel [21]

Answer:

A)

O = 3S

While O is the number of ounces of meat

While S is the number of sandwiches

B)

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