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Tatiana [17]
3 years ago
8

In the diagram of circle O, m NQM  103

Mathematics
1 answer:
-BARSIC- [3]3 years ago
5 0

Answer:

88 degree

Step-by-step explanation:

We assume the measure of MN is x degree.

As the measure of LP is 30 degree more than that of MN, so that the measure of LP is: x + 30 degree

In the circle, as 4 points M,N,P,L are on the circle, we have:

+) ∡MPN = 1/2 * (measure of ∡MPQMN) = x/2 = ∡MPQ

+) ∡LMP =1/2 * (measure of LP) = (x+30)/2 = ∡QMP

We have ∡NQM and ∡MQP are complementary angles, so that:

∡MQP + ∡NQM = 180

=> ∡MQP = 180 - ∡NQM = 180 -103 = 77

In the triangle QMP, total measure of 3 internal angles are 180 degree, so that:

∡MQP + ∡QMP + ∡MPQ = 180

=> 77 + (x + 30)/2 + x/2 = 180

=> 77 + x/2 + 15 + x/2 = 180

=> x = 180 -77-15= 88

So that the measure of MN is 88 degree

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3 years ago
A page should have perimeter of 42 inches. The printing area within the page
Tamiku [17]

Overall  dimensions of the page in order to maximize the printing area is  page should be 11 inches wide and 10 inches long .

<u>Step-by-step explanation:</u>

We have , A page should have perimeter of 42 inches. The printing area within the page  would be determined by top and bottom margins of 1 inch from each side, and the  left and right margins of 1.5 inches from each side. let's assume  width of the page be x inches  and its length be y inches So,

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width of printed area = x-3  & length of printed area = y-2:

area = length(width)

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Let's find \frac{d(area)}{dx}:

\frac{d(area)}{dx} = \frac{d(-x^{2}+22x-57)}{dx} = -2x +22 , for area to be maximum \frac{d(area)}{dx}= 0

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7 0
3 years ago
Hi! can you please help me and show me how you got your answer? thank you!
Kamila [148]

Answer:

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5 0
3 years ago
What is the slope of the line that passes through the points (1, 1) and (4, 4)?
gizmo_the_mogwai [7]

Answer:

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Step-by-step explanation:

Formula for finding the slope: \frac{y2-y1}{x2-x1}

Take y2 and y1 from (4,<u>4</u>) (1,<u>1</u>)

Take x2 and x1 from (<u>4</u>,4) (<u>1</u>,1)

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8 0
2 years ago
Help me with this... ​
marin [14]

i. 171

ii. 162

iii. 297

Solution,

n(U)= 630

n(I)= 333

n(T)= 168

i. Let n(I intersection T ) be X

333 - x + x + 468 - x = 630 \\ or \: 333 + 468 - x = 630 \\ or \: 801 - x = 630 \\ or \:  - x = 630 - 801 \\ or \:  - x =  - 171 \\ x = 171

<h3>ii.n(only I)= n(I) - n(I intersection T)</h3><h3> = 333 - 171</h3><h3> = 162</h3>

<h3>iii. n ( only T)= n( T) - n( I intersection T)</h3><h3> = 468 - 171</h3><h3> = 297</h3>

<h3>Venn- diagram is shown in the attached picture.</h3>

Hope this helps...

Good luck on your assignment...

4 0
4 years ago
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