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damaskus [11]
3 years ago
13

Hello can someone explain to me how to do this?

Mathematics
1 answer:
wel3 years ago
3 0
A)
you square both sides
by squaring the left side, you get rid of the square root leaving only x one that side. But you also need to square the other side because whatever you do to one side you have to do to the other.

Answer: x=o/a^2

b)
To get o on its own on one side you multiply a. You basically do the opposite of what a originally does to cancel it out. So originally, a is dividing o so if you do the opposite, a will multiply on both sides.

a X o cancels out the a, leaving only o.
But whatever you do to one side, you have to do to the other. so Vx (square root of x) has to also be multiplied by a.

Answer: Vx X a=o

c) this is basically the same as b).

apply what i said in b) to this and you will get your answer

to cancel out o, you multiply both sides by o.
because o is getting divided, you will have to do the opposite which is multiply o.

Answer: Vx X o=a





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The number of rows in the arena is 26

<h3>How to determine the number of rows?</h3>

The hockey arena illustrates an arithmetic sequence, and it has the following parameters:

  • First term, a = 220
  • Sum of terms, Sn = 10920
  • Common difference, d = 16

The number of rows (i.e. the number of terms) is calculated using:

S_n = \frac{n}{2}(2a + (n -1) * d)

So,we have:

10920 = \frac{n}{2}(2 * 220 + (n -1) * 16)

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21840 = n(440 + 16n -16)

Evaluate the like terms

21840 = n(424+ 16n)

Expand

21840 = 424n + 16n^2

Rewrite as:

16n^2 + 424n - 21840 = 0

Using a graphical tool, we have:

n = 26

Hence, the number of rows in the arena is 26

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brainly.com/question/6561461

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Find the value of x in each case: Given: Iso. ΔABC, HM ∥DG Find: x, m∠CAB, m∠CBA
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1. Start with ΔCIJ.

  • ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
  • the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
  • ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.

2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So

m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.

3. Consider ΔCKL.

  • ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
  • ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
  • the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
  • ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.

4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So

m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.

5. ΔABC is isosceles, then angles adjacent to the base are congruent:

m∠KBA=m∠JAB → 222°-8x=205°-7x,

7x-8x=205°-222°,

-x=-17°,

x=17°.

Then m∠CAB=m∠CBA=205°-7x=86°.

Answer: 86°.

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