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Katyanochek1 [597]
3 years ago
12

Hello can somedy help me please thank you in advance

Mathematics
1 answer:
alina1380 [7]3 years ago
4 0

Answer:

Area of the circle=πr²

= 3.142 \times  {4}^{2}  \\  = 50.272cm \\  = 50.3cm

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vichka [17]
-5|3+4k|=-115

start by dividing both sided by -5

\frac{-5}{-5}|3+4k|=-\frac{115}{-5}|3+4k|=23

when we have an absolute we obtain two equations, one for the positive value and one for the negative value

\begin{gathered} 3+4k=23 \\ -(3+4k)=23 \end{gathered}

solve each of them

\begin{gathered} 3+4k=23 \\ 4k=23-3 \\ 4k=20 \\ k=\frac{20}{4}=5 \end{gathered}

for the second equation

\begin{gathered} -(3+4k)=23 \\ -3-4k=23 \\ -4k=23+3 \\ -4k=26 \\ k=-\frac{26}{4}=-6.5 \end{gathered}

the solutions for k are -6.5 and 5

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4 years ago
Pls help will give brainiest
MAXImum [283]

Answer:

113

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Someone please help with this question
vlada-n [284]

Answer:

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7 0
2 years ago
A uniform border on a framed photograph has the same area as the photograph. What are the outside dimensions of the border if th
Black_prince [1.1K]

Given dimension of photograph = 25cm * 20cm

So area of photograph = 500cm²

Now, let the width of frame border be = x

So, dimensions of photograph + frame= (25 + 2x)*(20 + 2x)

As given, the area of frame is same as photograph , so

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4x^{2}+90x-500=0

2x^{2}+45x-250=0

Solving this quadratic equation,

For the quadratic equation of the form ax^{2}+bx+c=0 , solution is:

x1,2 = \frac{-b+\sqrt{b^{2}-4ac}}{2a}

Now putting a=2, b=45, c= -250 and solving he above equation, we get two values of x.

x = -27.11 (discard it as it is negative)   and

x = 4.61 cm

Hence the dimensions of the frame are:

20+2x = 20+2(4.61) = 29.22 cm

25+2x = 25+2(4.61) = 34.22 cm



7 0
3 years ago
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