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3 years ago

Doug’s pet gerbil eats 1/8 pound of food a day doug buys a 3/2-pound bag of food for his pet how many days will the food last

Mathematics

1 answer:

stealth61 [152]3 years ago

7 0

The food will last him 12 days

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If the scale factor of a dilation is 4/3 the image will be a reduction of the pre-image, because 4/3 is a fraction less than one

Law Incorporation [45]

True and sorry if I’m incorrect

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2 years ago

6. Identify the recursive formula for the sequence 3, –15, 75, –375, . . .

vlada-n [284]

Answer:

first option

Step-by-step explanation:

There is a common ratio r between consecutive terms in the sequence, that is

r = - 15 ÷ 3 = 75 ÷ - 15 = - 375 ÷ 75 = - 5

The recursive formula allows a term in the sequence to be found by multiplying the previous term by r , thus

f(n) = - 5f(n - 1) if n > 1

with f(1) = 3 ← first term

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3 years ago

If a human face only have 2 eye balls how many eye balls can rap there whole head? 1-10 in why?

IceJOKER [234]

22 I hope this helps

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3 years ago

Edgar owns 234 shares of Cawh Consolidated Bank, which he bought for $21.38 apiece. Each share pays a yearly dividend of $3.15.

trapecia [35]

Answer: B. The stocks have a yield 6.84 percentage points greater than that of the bonds.

Step-by-step explanation:

Firstly, the yield for stocks will be calculated as:

= return/ investment cost

= $3.15/$ 21.38

= 0.14733395

= 14.73%

The yield for bonds will be calculated as:

= Return/Investment cost

Return = 1,000 x 8.3% = 83

Investment cost = 1,000 x 105.166/100 = 1051.66‬

Yield = 83/1051.66

= 0.07892284

= 7.89%

Then, the difference between the yield will be:

= 14.73% - 7.89%

= 6.84%

Therefore, the stocks have a yield 6.84 percentage points greater than that of the bonds.

8 0

2 years ago

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

7 0

3 years ago