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2 years ago

Solve 8 = 2^x + 4 x = −4 x = −1 x = 0 x = 7

Mathematics

1 answer:

TEA [102]2 years ago

8 0

None of these answers match up to the equation. The answer is 2. 2^2, 2x2=4
4+4=8
8=2^2+4

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a baseball team won two out of their last four games. in how many different orders could they have two wins and two losses in fo

iren [92.7K]

$2 \times 3 = 6 \times 2 = 12$
12

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3 years ago

An insured's roof cost $4,000 when installed 5 years ago. It has been damaged by hail and must be replaced. The new roof will co

hoa [83]

Answer:

ACV=$4,500

Step-by-step explanation:

We have that the actual cash value (ACV) is defined as:

$ACV=\dfrac{R\times(E-C)}{E}$

Where:

$ACV =$ actual cash value

$R =$ replacement cost or purchase price of the item

$E =$ expected life of the item

$C =$ current life of the item

Then we have R=$6,000, C=5years, and to find the expected life of the item we can use the depreciating of the roof, then if the roof is depreciating $200 each year we just need to divide $4,000 by $200 to find the expected life of the roof:

$\dfrac{4,000}{200}=20$

Then the espected life of the roof is 20 years, with this result we have all the data, then:

$ACV=\dfrac{\$6,000\times (20-5)}{20}=\dfrac{\$6,000\times (15)}{20}=\dfrac{\$90,000}{20}=\$4,500$

Then the ACV is $4,500

5 0

3 years ago

What three that are equivalent to 4:3.

stich3 [128]

Three that are equivalent are 12:9 8:6 28:21

5 0

3 years ago

Factor 7x3y4−7x2y5 completely.

Anastasy [175]

Answer:

Step-by-step explanation:

when we factor completely, we take a look at what is the biggest common factor of 7x³y⁴ and 7x²y⁵. In this case it is 7x²y⁴.

7x³y⁴-7x²y⁵ = 7x²y⁴(x-y) which is B.

7 0

3 years ago

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

2 years ago