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3 years ago

5

If two lines have the same slope, they are parallel. True of false

1 answer:

kaheart [24]3 years ago

4 0

This is false. I believe the word you are looking for is "congruent".

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PLS help ASAP DUE IN 20 MINUTES I CANT FIGURE IT OUT

uysha [10]

Answer:

x=8

Step-by-step explanation:

if you multiply 6 by 8 and subtracted by 3 it will equal 45

3 0

3 years ago

Food costs are expected to rise 6% each month for the next year. Which series correctly depicts the cost (to the nearest cent) f

alexira [117]

For this case we have the following equation:
y = 150 * (1.06) ^ t
For the first month we have:
y = 150 * (1.06) ^ 1
y = 159 $
For the second month we have:
y = 150 * (1.06) ^ 2
y = 168.54 $
For the third month we have:
y = 150 * (1.06) ^ 1
y = 178.65 $
Answer:
d. $ 159.00 + $ 168.54 + $ 178.65

4 0

3 years ago

Read 2 more answers

Show that 0-58200-48826-5 is a valid UPC number

trasher [3.6K]

0-58200-48826-5djdjbdnd

3 0

3 years ago

Answer the question below

Delicious77 [7]

Answer:

a) The modal average is 3 bedrooms

b) The mean number of bedrooms is 2.825, or 3 rounded.

Step-by-step explanation:

If you lay out all the numbers from lowest to highest, and find the one in the middle, it is 3 bedrooms.

For the mean, if you add up all the number of bedrooms then divide by the total frequency, you get 2.825. This would be 113 ÷ 40.

3 0

3 years ago

If the price is increasing at a rate of 2 dollars per month when the price is 10 dollars, find the rate of change of the demand.

densk [106]

Answer:

The demand reduces by $7.12 per month

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Step-by-step explanation:

Given

$p\to price$

$x \to demand$

$2x^2+5xp+50p^2=24800.$

$p =10; \frac{dp}{dt} = 2$

Required

Determine the rate of change of demand

We have:

$2x^2+5xp+50p^2=24800.$

Differentiate with respect to time

$4x\frac{dx}{dt} + 5x\frac{dp}{dt} + 5p\frac{dx}{dt} + 100p\frac{dp}{dt} = 0$

Collect like terms

$4x\frac{dx}{dt} + 5p\frac{dx}{dt} = -5x\frac{dp}{dt} - 100p\frac{dp}{dt}$

Factorize

$\frac{dx}{dt}(4x + 5p) = -5(x + 20p)\frac{dp}{dt}$

Solve for dx/dt

$\frac{dx}{dt} = -\frac{5(x + 20p)}{4x + 5p}\cdot \frac{dp}{dt}$

Given that: $2x^2+5xp+50p^2=24800.$ and $p = 10$

Solve for x

$2x^2 + 5x * 10 + 50 * 10^2 = 24800$

$2x^2 + 50x + 5000 = 24800$

Equate to 0

$2x^2 + 50x + 5000 - 24800 =0$

$2x^2 + 50x -19800 =0$

Using a quadratic calculator, we have:

$x \approx -113\ and\ x\approx88$

Demand must be greater than 0;

So: $x=88$

So, we have: $x=88$ ; $p =10; \frac{dp}{dt} = 2$

The rate of change of demand is:

$\frac{dx}{dt} = -\frac{5(88 + 20*10)}{4*88 + 5*10} * 2$

$\frac{dx}{dt} = -\frac{5(288)}{402} * 2$

$\frac{dx}{dt} = -\frac{2880}{402}$

$\frac{dx}{dt} \approx -7.16$

<em>This implies that the demand reduces by $7.12 per month</em>

8 0

3 years ago