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ipn [44]
1 year ago
14

Quadrilateral SQUA is a square with sides measuring 25mm. Quadrilateral AQER is also a square.What is the length diagonal RQ

Mathematics
1 answer:
svet-max [94.6K]1 year ago
6 0

The square has 4 sides of the same measure.

If square SQUA has a side measuring 25 mm, it means side QU measures the same.

Also, if quadrilateral AQER is a square two, then its diagonals bisect each other, then QU is equal to UR.

Thus, QR measures:

\begin{gathered} QR=QU+UR \\ QR=25\operatorname{mm}+25\operatorname{mm} \\ QR=50\operatorname{mm} \end{gathered}

The answer is 50 mm

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Which of the following phrases are equations?
Svetradugi [14.3K]

Answer:

(A)

(B)

(D)

Step-by-step explanation:

(A) 60/5 = 4 × 3

(B) 74 = -2t

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Are equations

6 0
3 years ago
I'm have some issues with the problem shown in the screenshot and would love some help.
KiRa [710]

The dimensions and volume of the largest box formed by the 18 in. by 35 in. cardboard are;

  • Width ≈ 8.89 in., length ≈ 24.89 in., height ≈ 4.55 in.

  • Maximum volume of the box is approximately 1048.6 in.³

<h3>How can the dimensions and volume of the box be calculated?</h3>

The given dimensions of the cardboard are;

Width = 18 inches

Length = 35 inches

Let <em>x </em>represent the side lengths of the cut squares, we have;

Width of the box formed = 18 - 2•x

Length of the box = 35 - 2•x

Height of the box = x

Volume, <em>V</em>, of the box is therefore;

V = (18 - 2•x) × (35 - 2•x) × x = 4•x³ - 106•x² + 630•x

By differentiation, at the extreme locations, we have;

\frac{d V }{dx}  =  \frac{d( 4 \cdot \:  {x}^{3}  - 106 \cdot \:  {x}^{2}  + 630\cdot \:  {x} )  }{dx} = 0

Which gives;

\frac{d V }{dx}  =12\cdot \:  {x}^{2}  -  212\cdot \:  {x} + 630 = 0

6•x² - 106•x + 315 = 0

x  =  \frac{ - 6 \pm \sqrt{106 ^2 - 4 \times 6 \times 315} }{2 \times 6}

Therefore;

x ≈ 4.55, or x ≈ -5.55

When x ≈ 4.55, we have;

V = 4•x³ - 106•x² + 630•x

Which gives;

V ≈ 1048.6

When x ≈ -5.55, we have;

V ≈ -7450.8

The dimensions of the box that gives the maximum volume are therefore;

  • Width ≈ 18 - 2×4.55 in. = 8.89 in.

  • Length of the box ≈ 35 - 2×4.55 in. = 24.89 in.

  • Height = x ≈ 4.55 in.

  • The maximum volume of the box, <em>V </em><em> </em>≈ 1048.6 in.³

Learn more about differentiation and integration here:

brainly.com/question/13058734

#SPJ1

7 0
1 year ago
The course grade in a statistics class is the average of the scores on five examinations. Suppose that a student's scores on the
nadya68 [22]

Answer:

84 is the highest possible course average

Step-by-step explanation:

Total number of examinations = 5

Average = sum of scores in each examination/total number of examinations

Let the score for the last examination be x.

Average = (66+78+94+83+x)/5 = y

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x = 5y -321

If y = 6, x = 5×6 -321 =-291.the student cannot score -291

If y = 80, x = 5×80 -321 =79.he can still score higher

If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.

If y= 100

The average cannot be 100 as student cannot score 179(maximum score is 100)

8 0
3 years ago
5% of is 36<br> A. 72<br> B. 720<br> C. 7.2<br> D. 72.0
Shtirlitz [24]

Answer:

b. 720

Step-by-step explanation:

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Find the common difference of the arithmetic sequence.
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Each time you are subtracting 1/9 to get the next term

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(1 8/9) - (1/9) = 1 7/9

which is why the answer is choice A
7 0
3 years ago
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