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Vaselesa [24]
3 years ago
11

Answer for 14+3y < 65

Mathematics
2 answers:
ivanzaharov [21]3 years ago
5 0
Hey there!

14 + 3y < 65

SOME people replace “<”, “>”, “≥”, or “≤” as an equal sign (=) to make it easier to solve

14 + 3y < 65
14 + 3y = 65

SUBTRACT by 14 on both of your sides
3y + 14 - 14 = 65 - 14
CANCEL out: 14 - 14 because that gives you 0
KEEP: 65 - 14 because it helps you solve for “y”

**65 - 14 = 51**

NEW EQUATION: 3y = 51

DIVIDE by 3 both of your sides
3y/3 = 51/3
CANCEL out: 3/3 because that gives you 1
KEEP: 51/3 because it gives you the value of “y”



NEW EQUATION: y = 51/3

**51/3 = 17**

Answer: y = 17 ☑️

☑️ OVERALL ANSWER FOR YOU: y < 17 ☑️
NOTE: it’s an OPEN circle SHADED to the LEFT

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

Anarel [89]3 years ago
3 0

Answer:

The answer is y<17

Step-by-step explanation:

First, move the constant to the right.

3y<65-14

Then, subtract 65 and 14

3y<51

Lastly, divide both sides by 3.

y<17

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9/3 *[(18-10)-2^2]<br><br><br>36<br><br>12<br><br>0.25<br><br>0.75
poizon [28]

Answer:

12

Step-by-step explanation:

Use the PEMDAS

there is a parenthesis and exponents

9/3{(8) - 4)}

Multiply and divide 9/3= 3

Also subtract 8-4 = 4

3{8-4}

3(4)= 12

5 0
3 years ago
Which value of n makes this equation true?
jonny [76]

Answer:

the answer is A. -16

Step-by-step explanation:

3(-16)+3= -45

-45/5=

-9

5(-16)-1=-81

-81/9=

-9

-9=-9

4 0
3 years ago
The area of a room is 600 square feet. The length is (x + 5) feet and the width is (x + 4) feet. Find the dimensions of the room
tangare [24]
I'm going to assume that the room is a rectangle.

The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle. 

You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
A = lw\\&#10;600 = (x+5)(x+4)\\&#10;600 =  x^{2} + 9x + 20&#10;

Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
600 = x^{2} + 9x + 20\\&#10;x^{2} + 9x - 580 = 0\\&#10;(x + 29)(x - 20) = 0\\&#10;x + 29 = 0, \:\: x - 20 = 0\\&#10;x = -29, x = 20

Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.

The dimensions of your room are 25ft (length) by 24ft (width).
8 0
3 years ago
Read 2 more answers
Please answer #1 and #2
DIA [1.3K]
1)
x^2 + (3y/2z) = 7
2x^2z + 3y = 14z
3y = 14z - 2x^2z
3y = 2z(7 - x^2)
  y = 2/3(z)(7 - x^2)

2)
(3zx^4) /(5+z) = 2y
3zx^4 = 2y(5+z)
3zx^4 = 10y + 2yz

6 0
3 years ago
Can you please help me
max2010maxim [7]
Here’s the answer and each step I took to solve it. With the example I made on the side just continue to repeat that for each one to get the answers. You just replace that one number for the X to take the place for the next X value as you go down the table solving them.

7 0
3 years ago
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