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3 years ago

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Mathematics

2 answers:

Klio2033 [76]3 years ago

7 0

Answer:

A. A rate with one in the numerator

Step-by-step explanation:

We can find the unit rate when given a rate by dividing the unit in the numerator by the quantity in the denominator. For example, if we are given the rate $4.50/5 gallons of milk, we can divide 4.50 by 5 to find the unit price, which is dollars per one gallon of milk. A unit rate means a rate for one of something. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate.

Mazyrski [523]3 years ago

5 0

Answer:

C. A rate with one (1) in the denominator

A unit rate is basically a ratio of something to one.

example:

if 10 bananas costs ten dollars then one banana would cost one dollar.

Step-by-step explanation:

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Explain Long division in your own words using at least 3 sentences.

Marina86 [1]

Answer:

form of devision that takes longer bc it a fraction or decimal. first you want to get rid of the decimals. then devide the numbers from the lowest to the highest

I may be wrong been for ever since I did it on paper

4 0

3 years ago

Read 2 more answers

A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals

Sophie [7]

Answer:

a) $w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

b) $202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

c)For this case $\epsilon = \pm 1$ since that's the tolerance 1C

d) $\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smalles value on this case $\delta =0.113$

e) For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

$T(w)= 0.1 w^2 +2.156 w +20$

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

$203= 0.1w^2 +2.156 w +20$

And we can rewrite the expression like this:

$0.1w^2 +2.156 w-183=$

And we can solve this using the quadratic formula given by:

$w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where a =0.1, b =2.156 and c=-183. If we replace we got:

$w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

$202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case $\epsilon = \pm 1$ since that's the tolerance 1C

Part d

For this case we can do this:

$\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smallest value on this case $\delta =0.113$

Part e

For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

8 0

4 years ago

HELP ME PLEASEEEE ASAP!!!!

Fed [463]

Answer:

dang I have no idea

Step-by-step explanation:

I don't have one ‍♂️

8 0

3 years ago

You are dealt two cards without replacement what is the probability the first card is one of 4 twos and second is 4 tens in a de

Ilia_Sergeevich [38]

A = drawing a two card
B = drawing a ten card

P(A) = 4/52 = 1/13
P(B|A) = P(A and B)/P(A)
P(A)*P(B|A) = P(A and B)
P(A and B) = P(A)*P(B|A)
P(A and B) = (1/13)*(4/51)
P(A and B) = 4/663
P(A and B) = 0.006033

The answer as a fraction is exactly 4/663
The answer in decimal form is approximately 0.006033

8 0

3 years ago

Need help on number 16. A recipe calls for 3 cups of sugar for every 5 cups of flour. If the recipe you make requires 2.5 cups o

levacccp [35]

Um what if you multiply 2.5 x5=12.5 then multiply 16x3=48 and then subtract 12.5 by 48 and you will get 77 cups of sugar srry if it is not the right answer this is not my account this is my fathers im not in collage im in 5th grade . so i hope this helped you

5 0

3 years ago