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mario62 [17]
4 years ago
9

At a local animal shelter there are 12 small - size dogs and 5 medium - size dogs. every day, the small - size dogs are each giv

en 12.5 ounces of dry food and the medium - size dogs are each given 18 ounces of the same dry food. how many pounds of dry food does the shelter serve in one day?
Mathematics
2 answers:
marusya05 [52]4 years ago
8 0
12 small dogs × 12.5 ounces = 150 ounces
5 medium dogs × 18 ounces = 90 ounces

150 plus 90 is equal to 240
and 240 ounces is coverted into 15 pounds.
Anastaziya [24]4 years ago
4 0
19.0625 pounds of dry food in one day
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Which one of these formulas describes the following sequence?<br><br> 1,5,12,22,35
kati45 [8]

We can find a formula for nth term of the given sequence as follows:

1, 5, 12, 22, 35

The 1st differences between terms:

4, 7, 10, 13

The 2nd differences :

3, 3, 3

Since it takes two rounds of differences to arrive at a constant difference between terms, the nth term will be a 2nd degree polynomial of the form: a n^2 + b n + c, where c is a constant. The coefficients a, b, and the constant c can be found.

We can form the following 3 equations with 3 unknowns a, b, c:

1 = a\cdot1^2 + b\cdot1 + c\\5 = \cdot2^2 + b\cdot2 + c\\12 = a\cdot3^2 + b\cdot3 + c

Solving for a, b, c, we get:

a = 3/2, b = -1/2, c = 0

Therefore, the nth term of the given sequence is:

\boxed{ a_n = \dfrac{3}{2}n^2- \dfrac{1}{2} n}

7 0
3 years ago
Krisi took out a $500 discounted loan calculated using a simple interest rate of 5% for a period of 2 years. How much money does
Rama09 [41]

Answer:

$455

Step-by-step explanation:

The computation of the principal is shown below:

As we know that

Amount = Simple interest + Principal

Let us assume the principal be X

So,

Amount = $500

Simple interest = Principal × rate of interest × time period

                         = X × 5% × 2 years

                         = X × 0.1

So, the principal is

$500 =  X × 0.1 + X

$500 = 1.1 × X

So, the X is $455 i.e principal

     

5 0
4 years ago
Read 2 more answers
A restaurant sells curried cashew soup. On Monday the cook uses 3/5 of a bag of cashew. On Tuesday the cook uses ½ as any cashew
Vadim26 [7]

Answer:

The restaurant cooked 3/10 as many cashews on Tuesday

Step-by-step explanation:

If the word problem says that they used 3/5 of a bag of cashews on Monday, and they used 1/2 AS MANY.

So "as many" gives us a clue to multiply.

3/5 * 1/2 = 3/10

So the restaurant cooks 3/10 bag of cashews on Tuesday

3 0
3 years ago
Hello good day <br>can you help me please:(​
iVinArrow [24]

Answer:

31. A

32. D

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4 0
2 years ago
Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(
andreev551 [17]

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}

Where:

x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval

a. Let's find h(3) and h(4) using the data provided by the problem:

h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f

The average velocity over the interval [3, 4] is :

v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s

b. Let's find h(3.5) using the data provided by the problem:

h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f

The average velocity over the interval [3, 3.5] is :

v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s

c. Let's find h(3.1) using the data provided by the problem:

h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f

The average velocity over the interval [3, 3.1] is :

v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s

d. Let's find h(3.01) using the data provided by the problem:

h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f

The average velocity over the interval [3, 3.01] is :

v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s

e. Let's find h(3.001) using the data provided by the problem:

h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f

v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s

7 0
3 years ago
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