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2 years ago

HELPP ME PLEASE I NEED HELP

Mathematics

1 answer:

timofeeve [1]2 years ago

3 0

Answer:

See below.

Step-by-step explanation:

The equation is this

A= 1/2bh

We plug in the equation and we get this

A= 1/2 (14)(24)

Let's do the math.

A= 1/2 (336)

A= 168

That's your answer. ✌️

I hope this helps!

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Distributive property to write an equivalent expression 9(9m+3t)

Klio2033 [76]

Answer:

Step-by-step explanation:

9(9m+3t) = 9 *9m + 9*3t

= 81m + 27t

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3 years ago

Lynne feeds her cat 65 g of cat food per day. About how many ounces of cat food will she feed her cat in 5 weeks? 1 oz≈28.3 g

Svet_ta [14]

Consumption in Grams would be 65 x 7 x 5 which 2,275 If we use the conversion rate of 28.3 grams to an ounce then the number of ounces will be found by completing this sum 2275 divided by 28.3 which is 80 Ounces when rounded down

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3 years ago

The exact same experiment was conducted 14 times. How many times should the results have been similar for them to be valid?A.8

NeTakaya

Answer:

I would have to say C.14

Step-by-step explanation:

Because if you are running the same exact emperiment with the same exact materials everytime then you should get the same outcome.

Hope this helps! : D

4 0

3 years ago

Please help 15 points

Gwar [14]

Answer:

46/100

Step-by-step explanation:

The decimal reaches a point into the hundreths.

kiniwih426

5 0

2 years ago

Read 2 more answers

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has i

bija089 [108]

Answer: $\bold{a)\ P(t)=P_o\cdot e^{t\cdot ln(2.7)}}$

b) 5314

c) ln 2.7

d) 4.6 hrs

Step-by-step explanation:

$P(t) = P_o\cdot e^{kt}\\\\\bullet \text{P(t) is the number of bacteria after t hours} \\\bullet P_o\text{ is the initial number of bacteria}\\\bullet \text{k is the rate of growth}\\\bullet \text{t is the time (in hours)}\\\\\\270=100\cdot e^{k(1)}\\2.7=e^k\\ln\ 2.7=\ln e^k\\\boxed{ln\ 2.7=k}\\\\\text{So the equation to find the number of bacteria is: }\boxed{P(t)=P_o\cdot e^{t\cdot ln(2.7)}}\\\\\\P(4)=100\cdot e^{4\cdot ln(2.7)}\\.\qquad =\boxed{5314}$

$10,000=100\cdot e^{t\cdot ln(2.7)}\\100=e^{t\cdot ln(2.7)}\\ln\ 100=ln\ e^{t\cdot ln(2.7)}\\ln\ 100=t\cdot ln(2.7)\\\dfrac{ln\ 100}{ln\ 2.7}=t\\\\\boxed{4.6=t}$

4 0

3 years ago