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olga nikolaevna [1]
2 years ago
7

The ice cream Palace received 3 gallons of strawberry ice cream, 5 pints of mocha ice cream, and 1 quart of vanilla ice cream to

day. How many pints of ice cream did they receive?
Mathematics
1 answer:
Alina [70]2 years ago
6 0
They recived 31 pints of ice cream
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Anyone know this one??
egoroff_w [7]
I’d say option 3
Good luck!
8 0
2 years ago
Which exponential expression is equivalent to the one below?
OverLord2011 [107]

Answer:

The answer is A.

Step-by-step explanation:

You have to elaborate it :

{(54 \times ( - 13))}^{66}

=  {(54)}^{66}  \times  {( - 13)}^{66}

4 0
2 years ago
What is the y-intercept of line MN?<br><br> What is the equation of MN written in standard form?
bagirrra123 [75]

Answer:

y-intercept of the line MN = 2

Standard form of the equation ⇒ x + y = 2

Step-by-step explanation:

Coordinates of the ends of a line MN → M(-3, 5) and N(2, 0)

Slope of a line = \frac{y_2-y_1}{x_2-x_1}

                        = \frac{5-0}{-3-2}

                        = -1

Equation of the line MN passing through (-3, 5) and slope = -1,

y - 5 = (-1)(x + 3)

y - 5 = -x - 3

y = -x + 2

This equation is in the y-intercept form,

y = mx + b

where m = slope of the line

b = y-intercept

Therefore, y-intercept of the line MN = 2

Equation in the standard form,

x + y = 2

8 0
2 years ago
Read 2 more answers
A contaminant is leaking into a lake at a rate of R(t) = 1400e0.06t gallons/h. Enzymes have been added to the lake that neutrali
ruslelena [56]

Answer:

Step-by-step explanation:

Rate of leakage, R(t) = 1400 e^0.06t gallons/h

fraction remains , S(t) = e^(-0.32t)

initial contaminant = 1000 gallon

gallons contaminant present after t hour is S(t) R(t)

G(t) = S(t) R(t)

G(t) = 1400 e^{0.06t}\times e^{-0.32t}

G(t) = 1400 e^{- 0.26t}

Put t = 18 hours

G(t) = 1400 e^{- 0.26\times 18}

Taking log on both the sides

ln G = ln 1400 - 0.26 x 18

ln G = 7.244 - 4.68

ln G = 2.564

G = 13 gallons

5 0
2 years ago
The decay of 192 mg of an isotope is given by A(t)= 192e-0.015t, where t is time in years. Find the amount left after 55 years.
statuscvo [17]

Answer:

This is a typical radioactive decay problem which uses the general form:

A = A0e^(-kt)

So, in the given equation, A0 = 192 and k = 0.015. We are to find the amount of substance left after t = 55 years. That would be represented by A. The solution is as follows:

A = 192e^(-0.015*55)

A = 84 mg

6 0
2 years ago
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