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Rasek [7]
2 years ago
7

1 of 6 The length of each side of a square is 6.1 cm rounded to 1 decimal place. Write the error interval for the perimeter, p ,

in the form a ≤ p < b .
Mathematics
1 answer:
arsen [322]2 years ago
8 0

The error interval of the perimeter is 6.1 cm ≤ p < 6.15 cm

<h3>How to determine the error of interval?</h3>

The perimeter is given as:

P = 6.1 cm to 1 decimal place

The smallest number that can be approximated to 6.1 is 6.05, while the largest number is less than 6.15

This means that the interval is:

6.1 cm ≤ p < 6.15

Read more about error interval at:

brainly.com/question/16810055

#SPJ1

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Harrizon [31]
I hope this helps you

5 0
3 years ago
What are the solutions to the system of equations? {y=2x2+6x−10 y=−x+5
inessss [21]
Answer: x = -5, 3/2 y = 10, 7/2

Explanation:

y = 2x^2 + 6x - 10 (1)
y = - x + 5 (2)

=> 2x^2 + 6x - 10 = - x + 5
2x^2 - 7x - 15 = 0
(2x - 3)(x + 5) = 0
=> 2x - 3 = 0
2x = 3
x = 3/2
=> x + 5 = 0
x = -5

According to (2):

y = -x + 5
y = -(-5) + 5
y = 5 + 5 = 10

y = -x + 5
y = -(3/2) + 5
y = -(3/2) + 10/2
y = 7/2

3 0
3 years ago
Read 2 more answers
Please answer all I need help easy
Leokris [45]
The answer k is n divided by 6
8 0
3 years ago
Read 2 more answers
3.) mr. smith’s class sold wrapping paper for $3.50 each and mr. davis’ class sold magazines for $2.75 each. together, the class
igor_vitrenko [27]

Answer:

The number of wrapping paper sold was 32 and the number of magazines sold was 40

Step-by-step explanation:

Let

x ----> the number of wrapping paper sold

y ----> the number of magazines sold

we know that

The classes sold 72 items

so

x+y=72 ----> equation A

The classes earned $222 for their school

so

3.50x+2.75y=222 ----> equation B

Solve the system of equations by graphing

Remember that the solution of the system of equations is the intersection point both graphs

using a graphing tool

The solution is (32,40)

see the attached figure

therefore

The number of wrapping paper sold was 32 and the number of magazines sold was 40

7 0
3 years ago
Find the area of the shaded region.f(x)=4x+3x2−x3,g(x)=0<br> The area is _____.
Tasya [4]

Answer:

The answer is " \bold{\frac{7}{2}}"

Step-by-step explanation:

Given value:

\to f(x)=4x+3x^2-x^3\\\\\to g(x)=0

Find:

area=?

calculation:

\ Area = \int_{-1}^{0} -(4x+3x^2-x^3) dx  + \int_{0}^{1} (4x+3x^2-x^3) dx  \\\\

        =  -(2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\=  -( 2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\= ( 2- 1 -\frac{1}{4}) + (2+1- \frac{1}{4})\\\\\\=  ( \frac{8-4-1}{4}) + (\frac{8+4-1}{4})\\\\=  ( \frac{3}{4}) + (\frac{11}{4})\\\\=   \frac{3}{4} + \frac{11}{4}\\\\= \frac{11+3}{4}\\\\= \frac{14}{4}\\\\= \frac{7}{2}\\\\

7 0
3 years ago
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