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3 years ago

Find the difference.

Mathematics

2 answers:

sukhopar [10]3 years ago

6 0

Answer:

-2x-4

Step-by-step explanation:

(-7x-7y-4)-(-5x-7y)=-7x-7y-4+5x+7y=-7x-4+5x=-2x-4

adoni [48]3 years ago

6 0

Answer:

-2x - 4

Step-by-step explanation:

(- 7x - 7y - 4) - (-5x - 7y) Remove both sets of brackets
-7x - 7y - 4 + 5x + 7y Notice the sign change in 6x and 7y Two minus signs multiplied together make a plus. Gather like terms.
-7x + 5x - 7y + 7y - 4 Simplify
-2x - 4 Note the ys cancel out.

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28 lunch; 15 % tip I need to show my workkkk

NeX [460]

28 x .15 = 4.2

28 + 4.2 = 32.2 <---- ANSWER

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4 years ago

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Pi High School ordered 40 science books. The next week, the school ordered 30 algebra books. The bill for the first order was $3

maria [59]

Answer:

Each algebra book costs $60.

Step-by-step explanation:

A=bill for algebra books; S=bill for science books=A+$360

S+A=$3960 Substitute for S

A+$360+A=$3960 Subtract $360 from each side.

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A=$1800 The order for 30 algebra books cost $1800.

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3 years ago

Find the value of x. :]]

Mekhanik [1.2K]

Answer:

144

Step-by-step explanation:

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3 years ago

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What is the slope of the line?? PLZ help

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3 years ago

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago