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Alexxx [7]
3 years ago
14

For 3 consecutive odd integers, the result of adding the smallest integer to four times the largest integer is 121. What are the

3 integer?
Mathematics
1 answer:
irakobra [83]3 years ago
4 0

Answer: 21, 23, 25

Step-by-step explanation:

Let the integers be y, y+2 and y+4.

From the question, we are informed that the result of adding the smallest integer to four times the largest integer is 121. This can be mathematically expressed as:

y + 4(y+4) = 121

y + 4y + 16 = 121

5y + 16 = 121

5y = 121-16

5y = 105

y = 105/5

y = 21

Since y = 21,

y + 2 = 21 + 2 = 23

y + 4 = 21 + 4 = 25

The integers are 21,23 and 25

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Answer:

The mass of the last screw added is 3.855 grams.

Step-by-step explanation:

if you take 84.81 and subtract the original mass you should get the weight that was added.

Hope this helps.

3 0
3 years ago
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The area of a square patio is 196 square feet how long is each side of the patio ​
bonufazy [111]

Answer:

14 ft

Step-by-step explanation:

Let x be the length of each side

  1. Turn it into an algebraic expression: x^{2} = 196  
  2. 14 × 14 = 196
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I hope this helps!

7 0
3 years ago
Kim purchased 6 small containers of potato salad and Seth purchased 2 extra
victus00 [196]

Answer:

the difference is that the 2 extra LARGE containers weigh bigger than the

6 small ones

Step-by-step explanation:

6 0
3 years ago
A number x divided by -1 is at least -4
Vinil7 [7]
A number x divided by -1 is at least -4.

x / -1 ≥ -4

Multiply both sides by -1.

x ≤ 4

Remember, when multiplying or dividing by a negative, you must change the sign's direction.

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7 0
3 years ago
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP
erik [133]

Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

-x^{2} +2x+10-x-2=0

-x^{2} +x+8=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

5 0
3 years ago
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