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Svetach [21]
2 years ago
12

What is adjacent??????

Mathematics
1 answer:
alina1380 [7]2 years ago
5 0
Next to or adjoining something else.
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Triangle abc is a right triangle if side ac=5 and side ab=10 what is the measure of side bc
Nastasia [14]
You use the pythagoream thereom

10squared = 5squared plus xsquared

square root of (100-25)

the root of 75 would be you answer.

Hope this helps!
6 0
3 years ago
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Find measure of angle egf<br> A.45<br> B.60<br> C.90<br> D.100
astra-53 [7]

Answer:

It looks like C is the correct answer xd

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3 years ago
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When the new computer lab opened there were 18 computers.by the end of firdt week there were 25 computers. Find the percent of c
aliina [53]

The number of computers increased by 38.9%

Step-by-step explanation:

Number of computers at start of week = Old value = 18

Number of computers by end of week = New value = 25

Change = New value - Old value

Change = 25-18 = +7

The positive sign indicates the increase.

Increase percent = \frac{Change}{Old\ value}*100

Increase\ percent=\frac{7}{18}*100\\\\Increase\ percent=\frac{700}{18}\\\\Increase\ percent= 38.88\%

Rounding off to nearest tenth

Increase percent = 38.9%

The number of computers increased by 38.9%

Keywords: difference, percentage

Learn more about percentages at:

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3 0
3 years ago
Show that if f(n) is O(g(n)) and d(n) is O(h(n)) then f(n) + d(n) is O(g(n) + h(n)).
GaryK [48]
f(n)\in\mathcal O(g(n)) is to say

|f(n)|\le M_1|g(n)|

for all n beyond some fixed n_1.

Similarly, d(n)\in\mathcal O(h(n)) is to say

|d(n)|\le M_2|h(n)|

for all n\ge n_2.

From this we can gather that

|f(n)+d(n)|\le|f(n)|+|d(n)|\le M_1|g(n)|+M_2|h(n)|\le M(|g(n)|+|h(n)|)

where M is the larger of the two values M_1 and M_2, or M=\max\{M_1,M_2\}. Then the last term is bounded above by

M(|g(n)|+|h(n)|)\le2M\max\{|g(n)|,|h(n)|\}

from which it follows that

f(n)+d(n)\in\mathcal O(\max\{g(n),h(n)\})
3 0
3 years ago
Ellen is making jewelry sets that contain a bracelet and a pair of earrings. Each bracelet uses 3 times as many beads as one ear
dangina [55]
<h2 /><h2><em>So</em><em> </em><em>there</em><em> </em><em>are</em><em> </em><em>a</em><em> </em><em>pair</em><em> </em><em>of </em><em>earrings</em><em> </em><em>and</em><em> </em><em>a</em><em> </em><em>Brac</em><em>elet</em><em>.</em><em> </em></h2>

<em>It's </em><em>given</em><em> </em><em>that</em><em> </em><em>the</em><em> </em><em>Brac</em><em>elet</em><em> </em><em>uses</em><em> </em><em>3</em><em> </em><em>times</em><em> </em><em>the</em><em> </em><em>num</em><em>ber</em><em> </em><em>of</em><em> </em><em>beads</em><em> </em><em>that's </em><em>used</em><em> </em><em>in </em><em>making</em><em> </em><em>a</em><em> </em><em>single</em><em> </em><em>earrin</em><em>g</em><em>.</em><em> </em>

<em>It's </em><em>also</em><em> </em><em>given</em><em> </em><em>that</em><em> </em><em>one</em><em> </em><em>single</em><em> </em><em>earing</em><em> </em><em>has</em><em> </em><em>1</em><em>3</em><em> </em><em>beads</em><em>.</em><em> </em><em>So</em><em> </em><em>a</em><em> </em><em>single</em><em> </em><em>brac</em><em>elet</em><em> </em><em>would</em><em> </em><em>have</em><em> </em><em>(</em><em>3</em><em>×</em><em>1</em><em>3</em><em>)</em><em> </em><em>beads </em><em>.</em><em>.</em><em>.</em><em>.</em><em> </em><em>and</em><em> </em><em>that's </em><em>equal</em><em> </em><em>to</em><em> </em><em>3</em><em>9</em><em>.</em><em> </em>

<em>Making</em><em> </em><em>a</em><em> </em><em>single</em><em> </em><em>set</em><em> </em><em>of </em><em>jewellery</em><em> </em><em>needs</em><em> </em><em>a</em><em> </em><em>pair</em><em> </em><em>of </em><em>earr</em><em>ings</em><em> </em><em>and</em><em> </em><em>a</em><em> </em><em>Bracelet</em><em>.</em><em> </em>

<em>So</em><em> </em><em>total</em><em> </em><em>nu</em><em>mber</em><em> </em><em>of </em><em>required</em><em> </em><em>beads</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>=</em><em> </em>

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3 0
3 years ago
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