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ohaa [14]
4 years ago
8

Hello I need help on it ​

Mathematics
2 answers:
expeople1 [14]4 years ago
7 0
It would be 200 :)))))
romanna [79]4 years ago
7 0

42145357556574455247788555

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The HCF and LCM of two numbers x and 126 are 24 and 840 respectively. Find the value of x.
Zigmanuir [339]

Answer:

x=160

Step-by-step explanation:

x×126=24×840

x=\frac{24 \times840}{126} =160

8 0
3 years ago
What is the sum of the first 30 terms in a(n)=4n+1
dimulka [17.4K]
Hello,

\sum _{i=1} ^{30}(a_i)=\sum _{i=1} ^{30}(4*i+1)\\

=4*\sum _{i=1} ^{30}(i) +\sum _{i=1} ^{30}(1)\\

=4* \dfrac{31*30}{2} + 30 =1860+30=1890
6 0
3 years ago
What is the greatest perfect square that is a factor of 1290
Keith_Richards [23]

Answer:

There is no greatest perfect square of 1290

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find the range of each function, given the domain
Olenka [21]
In #12, substitute into the given function the x values {-3, 0, 6} to obtain the range:  the function takes on the following values:  {0, -1, -3}.  Thus, the range for #13 is {0, -1, -3}.

Do the others in precisely the same way.

7 0
4 years ago
I need help with this problem. Given: BC = 10 inches AC = √50 inches m∠CBD = 60° m∠CAD = 90° Calculate the exact area of the sha
Nikitich [7]

Answer:

Area of the shaded region = 23.33 in²

Step-by-step explanation:

Area of a sector = \frac{\theta}{360}(\pi r^{2})

Where θ = Central angle subtended by an arc

r = radius of the circle

Area of the sector BCD = \frac{60}{360}(\pi) (10^{2})

                                       = 52.36 in²

Area of equilateral triangle BCD = \frac{\sqrt{3} }{4}(\text{Side})^2

                                                      = \frac{\sqrt{3} }{4}(10)^2

                                                      = 25\sqrt{3} in²

                                                      = 43.30 in²

Area of the shaded portion in ΔBCD = 52.36 - 43.3

                                                             = 9.06 in²

Area of sector CAD = \frac{90}{360}(\pi)(\sqrt{50})^2

                                 = 39.27 in²

Area of right triangle CAD = \frac{1}{2}(\text{Base})(\text{Height})

                                            = \frac{1}{2}(\text{AC})(\text{AD})

                                            = \frac{1}{2}(\sqrt{50})(\sqrt{50})

                                            = 25 in²

Area of the shaded part in the ΔACD = 39.27 - 25

                                                                         = 14.27 in²

Area of the shaded part of the figure = 9.06 + 14.27

                                                                = 23.33 in²

8 0
3 years ago
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