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mamaluj [8]
3 years ago
12

Can someone help me with this word problem.

Mathematics
2 answers:
vaieri [72.5K]3 years ago
7 0
Yeah so the notebooks=n and binders=b

3n+7b=47
12+7b=47
-12 -12
7b= 35
b= 5
Sunny_sXe [5.5K]3 years ago
7 0
4 x 3 = 12

47-12 = 35

35 divided by 7 is 5

so I believe he bought 5 binders

if you add it up it like 5 x 7= 35+12= 47
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1. Which point on the axis satisfies the inequality y
grigory [225]

Answer:

1) Point (1,0) -----> see the attached figure N 1

2) The value of x is 4

3) I quadrant

4) (1,1)

5)  y>-5x+3

Step-by-step explanation:

Part 1)

we know that

If the point satisfy the inequality

then

the point must be included in the shaded area

The point (1,0) is included in the shaded area

Part 2)

we have

x-2y\geq 4

see the attached figure N 2

we know that

The value for x on the boundary line and the x axis is equal to the x-intercept of the line x-2y= 4

For y=0

Find the value of x

x-2(0)= 4  

x=4

The solution is x=4

Part 3)

we have

x\geq 0 -----> inequality A

The solution of the inequality A is in the first and fourth quadrant

y\geq 0 -----> inequality B

The solution of the inequality B is in the first and second quadrant

so

the solution of the inequality A and the inequality B is the first quadrant

Part 4) Which ordered pair is a solution of the inequality?

we have

y\geq 4x-5

we know that

If a ordered pair is a solution  of the inequality

then

the ordered pair must be satisfy the inequality

we're going to verify all the cases

<u>case A)</u> point (3,4)

Substitute the value of x and y in the inequality

x=3,y=4

4\geq 4(3)-5

4\geq 7 ------> is not true

therefore

the point (3,4) is not a solution of the inequality

<u>case B)</u> point (2,1)

Substitute the value of x and y in the inequality

x=2,y=1

1\geq 4(2)-5

1\geq 3 ------> is not true

therefore

the point  (2,1) is not a solution of the inequality

<u>case C)</u> point (3,0)

Substitute the value of x and y in the inequality

x=3,y=0

0\geq 4(3)-5

0\geq 7 ------> is not true

therefore

the point  (3,0) is not a solution of the inequality

<u>case D)</u> point (1,1)

Substitute the value of x and y in the inequality

x=1,y=1

1\geq 4(1)-5

1\geq -1 ------> is true

therefore

the point  (1,1) is  a solution of the inequality

Part 5) Write an inequality to match the graph

we know that

The equation of the line has a negative slope

The y-intercept is the point (3,0)

The x-intercept is a positive number

The solution is the shaded area above the dashed line

so

the equation of the line is y=-5x+3

The inequality is  y>-5x+3

3 0
3 years ago
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I need help with a question please help me: You have to write just the equation i can do the rest
Sholpan [36]

Answer:

Step-by-step explanation:

x+2×3x=26

5 0
3 years ago
Given limit f(x) = 4 as x approaches 0. What is limit 1/4[f(x)]^4 as x approaches 0?
stepladder [879]

Answer:

\displaystyle 64

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Limit Rule [Variable Direct Substitution Exponential]:                                         \displaystyle \lim_{x \to c} x^n = c^n

Limit Property [Multiplied Constant]:                                                                     \displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle  \lim_{x \to 0} f(x) = 4

<u>Step 2: Solve</u>

  1. Rewrite [Limit Property - Multiplied Constant]:                                           \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4} \lim_{x \to 0} [f(x)]^4
  2. Evaluate limit [Limit Rule - Variable Direct Substitution Exponential]:       \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4}(4^4)
  3. Simplify:                                                                                                         \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = 64

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Book: College Calculus 10e

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2 years ago
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jarptica [38.1K]
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3 years ago
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The Answer:

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