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almond37 [142]
3 years ago
14

Connie's puppy Fido weighs 12 pounds. The veterinarian says that Fido will gain weight at a steady rate for the next 6 months an

d weigh at least 27 pounds after that. What inequality can you write to find the average number of pounds the veterinarian expects Fido to gain monthly for the next six months?
Mathematics
2 answers:
makvit [3.9K]3 years ago
8 0

Answer:42

Step-by-step explanation:

Tamiku [17]3 years ago
6 0

Answer:

6x + 12 \geq 27

Step-by-step explanation:

x is defined as the average number of pounds Fido will gain each month. You multiply x by 6, as this is the amount of months you are calculating for, to get the amount of weight that Fido will gain in the next 6 months. You add this value to 12, as this was Fido's starting weight. You use the "greater than or equal to" sign because he is going to weigh at least 27 pounds, which means 27 at the minimum, and potentially more.

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The bigger of two numbers is four more
Otrada [13]

Answer:

X=the smaller

y=the larger

The bigger of two numbers is four more than the smaller, then:

y=x+4

one more than twice the smaller is the same as the larger, then

1+2x=y

We have the following system of equations:

y=x+4

y=2x+1

We solve this system by equal values method:

x+4=2x+1

x-2x=1-4

-x=-3

x=3

We find the value of "y"now:

y=x+4

y=3+4

y=7

the values of x and y are:

x=3

y=7

Step-by-step explanation:

7 0
3 years ago
Identify the effect on the graph of replacing f(x) by f(x - h)
Tresset [83]

Given:

Replace f(x) by f(x - h).

To find:

The effect on the graph of replacing f(x) by f(x - h).

Solution:

Horizontal shift is defined as:

If the graph f(x) shifts h units left, then f(x+h).

If the graph f(x) shifts h units right, then f(x-h).

Where, h is a constant that represents the horizontal shift.

In the given problem f(x) is replaced by f(x - h) and we need to find the effect on the graph.

Here, we have x-h in place of x.

Therefore, the graph of f(x) shifts h units right to get the graph of f(x-h).

4 0
3 years ago
Two point particles are attached to each other via a light in-extensible string that passes over a ring hanging from the ceiling
arlik [135]

a. The expression for the tension in the string is T = mg

b. The expression for the tension in the chain is T = 2mg

To solve the problem, we need to know what tension is.

<h3>What is tension?</h3>

This is the opposing force in a stretched material.

<h3>a. Tension in the string</h3>

The expression for the tension in the string is T = mg

Since the string supports the weight of the object, the tension in the string equals the weight of the particle.

Let

  • T = tension in sring and
  • W = weight of particle = mg where
  • m = mass of particle and
  • g = acceleration due to gravity

So, T = W

T = mg

So, the expression for the tension in the string is T = mg

<h3>b. Tension in chain</h3>

The expression for the tension in the chain is T = 2mg

<h3 />

Since the chain supports both strings, the tension in the chain equals the tension in both strings.

Let

  • T' = tension in chain and
  • T = tension in each string

So, T' = T + T

T' = 2T

T' = 2W

T' = 2mg

So, the expression for the tension in the chain is T = 2mg

<h3 />

Learn more about tension here:

brainly.com/question/24994188

5 0
2 years ago
Solve the equation C^2 =4
matrenka [14]

Answer:

2

Step-by-step explanation:

Hey there!

The equation is asking us, c x c = 4

What is c?

Well the only number that is equal to 4 when you square it is 2

8 0
1 year ago
a car was valued at $41,000 in the year 2009 by 2013 the car value has depreciated to 19,000 if the car value continues to by th
lubasha [3.4K]

Answer:

$6,376.92

Step-by-step explanation:

-Let d be the rate of depreciation per year.

-Therefore, the value after n years can be expressed as:

A=P(1-d)^n\\\\A=Value \ after \ n  \ years\\P=Initial \ Value\\d=Rate \ of \ depreciation\\n=Time \ in \ years

#We substitute for the years 2009-2013 to solve for d:

A=P(1-d)^n\\\\19000=41000(1-d)^4\\\\0.475=(1-d)^4\\\\d=1-0.475^{0.25}\\\\d=0.1698

#We then use the calculated depreciation rate above to solve for A after 10 yrs:

A=P(1-d)^n\\\\=41000(1-0.1698)^{10}\\\\=\$6,376.92

Hence, the value of the car after 10 yrs is $6,376.92

3 0
3 years ago
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