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2 years ago

Solve for x. -15x + 60 ≤ 105 AND 14x + 11 ≤ -31

Mathematics

2 answers:

givi [52]2 years ago

7 0

Answer:

x ≥ -3
x ≥ -3

Step-by-step explanation:

-15x + 60 ≤ 105

-15x ≤ 45

x ≥ -3

14x + 11 ≤ -31

14x ≤ -42

x ≥ -3

Best of Luck!

Nataliya [291]2 years ago

7 0

Answer:

x=-3

Step-by-step explanation:

I just took it

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A study shows that, over the course of a lifetime, the average salary of college graduates is greater than the average salary of

iogann1982 [59]

Answer: The word "ensures" implies that it is certain.

In the case of the study, they simply found that the average salary is greater for college graduates. This doesn't mean that in every case the college graduate makes more money.

The problem is the word "ensures". Our study doesn't show certainty only a general pattern.

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2 years ago

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Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

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2 years ago

anna can buy 3 sweatshirts for a total of $45.how much would it cost if she were to buy 5 sweatshirts at same price?

Feliz [49]

Answer: Anna could buy 5 sweatshirts for $75.

Step-by-step explanation:

lets figure out the price of one sweatshirt.

45/3 = 15

one sweatshirt is $15

5*15=75

so 5 sweatshirts cost $75

6 0

2 years ago

A baker determined the annual profit in dollars from selling pies using p n( ) = 52n − 0.05n2 , where n is the number of pies so

miss Akunina [59]

Answer:

<u>The annual profit if the baker sells 400 pies would be US$ 12,800</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Annual profit in dollars from selling pies using p n( ) = 52n − 0.05n² , where n is the number of pies sold

Annual sales of the baker = 400 pies

2. Let's solve p for 400 pies

p n( ) = 52n − 0.05n²

p (n₄₀₀) = 52 * 400 − 0.05 * 400²

p (n₄₀₀) = 20,800 − 0.05 * 160,000

p (n₄₀₀) = 20,800 − 8,000

p (n₄₀₀) = 12,800

<u>The annual profit if the baker sells 400 pies would be US$ 12,800</u>

Note: Same answer to question 14179960, answered by me today.

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2 years ago

A trust fund worth $25,000 is invested in two different portfolios, this year, one portfolio is expected to earn 5.25% interest

OLEGan [10]

Answer:

The amount of money that should be invested at the rate of 5.25% is $12,000 and the amount money that should be invested at the rate of 4% is $13,000

Step-by-step explanation:

we know that

The simple interest formula is equal to

$I=P(rt)$

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

Let

x ------> the amount of money that should be invested at the rate of 5.25%

25,000-x -----> the amount money that should be invested at the rate of 4%

in this problem we have

$t=1\ year\\ P_1=\$x\\P_2=\$(25,000-x)\\r_1=0.0525\\r_2=0.04\\I=\$1,150$

substitute in the formula above

$I=P_1(r_1t)+P_2(r_2t)$

$1,150=x(0.0525*1)+(25,000-x)(0.04*1)$

Solve for x

$1,150=0.0525x+1,000-0.04x$

$0.0525x-0.04x=1,150-1,000$

$0.0125x=150$

$x=\$12,000$

$|$25,000-x=\$13,000$

therefore

The amount of money that should be invested at the rate of 5.25% is $12,000 and the amount money that should be invested at the rate of 4% is $13,000

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2 years ago

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