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jonny [76]
3 years ago
7

3 lb 5 oz X 6 = __lb __ oz

Mathematics
1 answer:
hoa [83]3 years ago
6 0
19lbs 14 ounces because 16 oz= 1lb
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Solve this quadratic equation by completing the square. x2 + 10x = 26
Misha Larkins [42]
                        x² + 10x = 26
                x² + 10x + 25 = 26 + 25
          x² + 5x + 5x + 25 = 51
x(x) + x(5) + 5(x) + 5(5) = 51
         x(x + 5) + 5(x + 5) = 51
                 (x + 5)(x + 5) = 51
                          (x + 5)² = 51
                             x + 5 = ±√(51)
                                 - 5      - 5
                                   x = -5 ± √(51)
                                   x = -5 + √(51)    or    x = -5 - √(51)
5 0
3 years ago
Read 2 more answers
Descompuneti in factori. x^2+4x-5=?
Lapatulllka [165]
<span>Assume that,
   x^2+4x-5 = 0 .......(1)
       Then, x^2+4x-5 = 0 x^2+5x-1x-5 =0
   x(x+5)-1(x+5) = 0
    (x+5) (x-1) = 0
       We get x=-5 and x=1
       Sub x=-5 in equ (1)
   (-5)^2+4(5)-5 = 0
     -25+20-5 = 0
    -25+25= 0
    0 = 0
       Sub x=1 in equ (1) (1)^2+4(1)-5 = 0
    1+4-5 = 0 5-5 = 0
     0 = 0
       Therefore x value is -5 and 1</span>
3 0
3 years ago
You need to find the LCM of 13 and 14. Would you rather use their multiples or their prime factorizations? Explain.
Vladimir [108]

Answer:

I would rather use their multiples.

Step-by-step explanation:

With 13 and 14, neither number has many factors (and few prime factors, for that matter) so factoring would be pretty much useless. If you wanted to use the prime factors of 14 (7 and 2) multiplying either of them by 13 would not give you the LCM. Actually the LCM is just 13*14 which is 182.

Good luck!

5 0
3 years ago
The figure is a circle with center O.
STatiana [176]

<u>Given</u>:

Given that the circle with center O.

The radius of the circle is OB.

The chord of the circle O is PQ and the length of PQ is 12 cm.

We need to determine the length of the segment PA.

<u>Length of the segment PA:</u>

We know that, "if a radius is perpendicular to the chord, then it bisects the chord and its arc".

Thus, we have;

PA=\frac{PQ}{2}

Substituting the value PQ = 12, we get;

PA=\frac{12}{2}

PA=6 \ cm

Thus, the length of the segment PA is 6 cm.

Hence, Option d is the correct answer.

3 0
3 years ago
A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman
Yanka [14]

Answer:

W\approx 8.72 and L\approx 15.57.

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

2L+W+\frac{1}{2}(2\pi r)=38

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

2L+W+\frac{1}{2}(2r\pi)=38

2L+W+\frac{\pi}{2}W=38

Let us find area of window equation as:

\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)

\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})

\text{Area}=W\cdot L+\frac{\pi}{8}W^2

Now, we will solve for L is terms W from perimeter equation as:

L=38-(W+\frac{\pi }{2}W)

Substitute this value in area equation:

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2

Since we need the area of window to maximize, so we need to optimize area equation.

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2  

A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2  

Let us find derivative of area equation as:

A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W  

A'=38-2W-\pi W+\frac{\pi}{4}W    

A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W

A'=38-2W-\frac{3\pi W}{4}

To find maxima, we will equate first derivative equal to 0 as:

38-2W-\frac{3\pi W}{4}=0

-2W-\frac{3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}*4=-38*4

-8W-3\pi W=-152

8W+3\pi W=152

W(8+3\pi)=152

W=\frac{152}{8+3\pi}

W=8.723210

W\approx 8.72

Upon substituting W=8.723210 in equation L=38-(W+\frac{\pi }{2}W), we will get:

L=38-(8.723210+\frac{\pi }{2}8.723210)

L=38-(8.723210+\frac{8.723210\pi }{2})

L=38-(8.723210+\frac{27.40477245}{2})

L=38-(8.723210+13.70238622)

L=38-(22.42559622)

L=15.57440378

L\approx 15.57

Therefore, the dimensions of the window that will maximize the area would be W\approx 8.72 and L\approx 15.57.

8 0
3 years ago
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