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Sliva [168]
3 years ago
10

A rectangle of sides 5 cm and 2 cm is cut out from a bigger rectangle of sides 14 cm and 7 cm to form a new shape as shown below

.
What is the perimeter of the new shape formed?

a. 46 cm
b. 40 cm
c. 22 cm
d. 20 cm

Mathematics
2 answers:
iren2701 [21]3 years ago
6 0
A. 46cm

7+7+14+2+2+5 + the remaining 9cm we get from 14-5cm
masha68 [24]3 years ago
3 0

Answer: a. 46 cm

Step-by-step explanation:

(7 + 14) × 2 + 2× 2 = 46 (cm)

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N
GrogVix [38]

9514 1404 393

Answer:

  (d)   f(x) = 2x^2 - 16x + 35

Step-by-step explanation:

The x-coordinate of the extreme will be found at ...

  x = -b/(2a)

where the function is f(x) = ax²+bx+c.

The extreme will be a minimum when a > 0. (eliminates choices A and B)

The x-coordinates of the extremes are ...

  C: -(-4)/(2(4)) = 1/2

  D: -(-16)/(2(2)) = 4 . . . . . matches the requirement

The appropriate choice is ...

  f(x) = 2x^2 - 16x + 35

3 0
3 years ago
The thicknesses of 81 randomly selected aluminum sheets were found to have a variance of 3.23. Construct the 98% confidence inte
Yuliya22 [10]

Answer:

The confidence interval for the population variance of the thicknesses of all aluminum sheets in this factory is Lower limit = 2.30, Upper limit = 4.83.

Step-by-step explanation:

The confidence interval for population variance is given as below:

[(n - 1)\times S^{2}  /  X^{2}  \alpha/2, n-1 ] < \alpha < [(n- 1)\times S^{2}  / X^{2} 1- \alpha/2, n- 1 ]

We are given

Confidence level = 98%

Sample size = n = 81

Degrees of freedom = n – 1 = 80

Sample Variance = S^2 = 3.23

X^{2}_{[\alpha/2, n - 1]}   = 112.3288\\\X^{2} _{1 -\alpha/2,n- 1} = 53.5401

(By using chi-square table)

[(n – 1)*S^2 / X^2 α/2, n– 1 ] < σ^2 < [(n – 1)*S^2 / X^2 1 -α/2, n– 1 ]

[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]

2.3004 < σ^2 < 4.8263

Lower limit = 2.30

Upper limit = 4.83.

8 0
3 years ago
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
Using number line what is 4-5​
Vinil7 [7]

Answer:

<h2>-1</h2>

Step-by-step explanation:

4 - 5 = -1

Count:

4, 3, 2, 1, 0, <u>-1</u>

6 0
3 years ago
{[14-2×3]-[18÷(12-3)]}+{42-11×3+[(65-9×5)+3]}
natali 33 [55]
Use Bidmas, solve all the inner brackets (), with first division, then multiplication then addition then subtraction
5 0
3 years ago
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