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2 years ago

11

So Alex is selling a very rare spice that costs $100 per gram. Someone wants to buy $80 worth of the spice. How much spice (in g

rams) does Alex give him?

2 answers:

Alika [10]2 years ago

7 0

80% of a gram.
One gram is $100, aka $100 is 100% of a gram, so $80 would be 80% of a gram.

svetlana [45]2 years ago

6 0

80% of one gram is 0.8 grams. It's literally that simple.

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Help please less than 19 minutes to turn in

vagabundo [1.1K]

Angle K = angle M

(4x-1)=(2x+27)

4x-2x=27+1

2x=28

x=14

K=4x-1 then 4(14)-1
55 degrees
M=2x+27 then 2(14) +27
55 degrees
K+M+L =180
55+55+L=180
110+L=180
L=70 degrees
Therefore K=55 degrees M=55 degrees L=70 degrees

3 0

3 years ago

Juliette made cookies for a bake sale and wants to package the same number of cookies together. If she makes 16 chocolate chip c

pentagon [3]

Answer:

8 cookies

Step-by-step explanation:

We are asked in the above question to find the most number of cookies that would be in the bag.

We solve the above question using the Greatest Common Factor method

We find the factors of 16, 24, 32

The factors of 16 are: 1, 2, 4, 8, 16

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

The factors of 32 are: 1, 2, 4, 8, 16, 32

Then the greatest common factor is 8.

Therefore, the most number of each type of cookie that will be in the bag is 8 cookies

4 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

Somebody help me with this question

vampirchik [111]

I think u do 3•5 then divide by 110

3 0

3 years ago

Read 2 more answers

Please help me answer this

yarga [219]

80,000going up by 10% you just add the 10% on

6 0

2 years ago