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Naddik [55]
3 years ago
11

Someone help me!!!please

Mathematics
1 answer:
gregori [183]3 years ago
8 0

9514 1404 393

Answer:

  115 square units

Step-by-step explanation:

The area of a parallelogram is the product of base length and height. The height is the perpendicular distance between the bases.

  A = bh

  A = (23)(5) = 115 . . . . square units

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A delivery person uses a service elevator to bring boxes of books up to an office. The delivery person weighs 190 lb and each bo
vagabundo [1.1K]

Answer:

25 boxes.

Step-by-step explanation:

We have the inequality:

190 + 60x ≤ 1740  where  x = the number of boxes of books.

60x  ≤  1740 - 190

60x   ≤  1550

x ≤ 25.83.

6 0
3 years ago
Read 2 more answers
Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
Answer this pls..................​
Sindrei [870]

Answer:

so we know:

total of 20 students

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

chemistry has a total of 14 students

1 2 3 4 5 6 7 8 9 10 11 12 13 14

8 students are in physics

1 2 3 4 5 6 7 8

3 are in all of them

1 2 3

1) underline the main 3

total of 20 students

<u>1 2 3 </u>4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

chemistry has a total of 14 students

<u>1 2 3</u> 4 5 6 7 8 9 10 11 12 13 14

8 students are in physics

<u>1 2 3 </u>4 5 6 7 8

2) lets bold the physics students, if you are in physics you have to do chem so that means 8 of the chem students also take physics.

total

<u>1 2 3 </u>4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

chemistry

<u>1 2 3</u> 4 5 6 7 8 9 10 11 12 13 14

physics

<u>1 2 3 </u>4 5 6 7 8

3) now we bring biology into the "equation"

off the bat we know 3 are definitely in Biology.

so if 4 study ONLY biology we can cross them off our list see the {}

total

<u>1 2 3 </u>4 5 6 7 8 {9 10 11 12} 13 14 15 16 17 18 19 20

chemistry

<u>1 2 3</u> 4 5 6 7 8 {9 10 11 12} 13 14

physics

<u>1 2 3 </u>4 5 6 7 8

biology

<u>1 2 3 </u>

<u />

4) so we have a total of 20 students, 3 in all of the classes <u>(underlined)</u>, 8 in physics (bolded). If there 14 in chemistry, 4 take ONLY chemistry {}, 8 take chemistry as well (bolded), that leaves two students, so they must belong in biology. lets<em> italicise</em> them.

total

<em><u>1 2 3 </u></em>4 5 6 7 8 {9 10 11 12} <em>13 14</em> 15 16 17 18 19 20

chemistry

<em><u>1 2 3</u></em><em> </em>4 5 6 7 8 {9 10 11 12} <em>13 14</em>

physics

<em><u>1 2 3</u></em><u> </u>4 5 6 7 8

biology

<em><u>1 2 3 </u></em><em> 13 14</em>

<u />

5) out of our total of 20 students we know 14 of the students classes, so the remaining six must be in biology.

<u>1 2 3 </u>4 5 6 7 8 {9 10 11 12} <em>13 14</em> 15 <em>16 17 18 19 20</em>

chemistry

<u>1 2 3</u> 4 5 6 7 8 {9 10 11 12} <em>13 14</em>

physics

<u>1 2 3 </u>4 5 6 7 8

biology

<u>1 2 3 </u><em>13 14</em> 15 <em>16 17 18 19 20</em>

6) so that was our "diagram" so now lets count how many are in chemistry and biology but NOT physics.

<u>1 2 3 </u>4 5 6 7 8 {9 10 11 12} <em>13 14</em> 15 <em>16 17 18 19 20</em>

chemistry

<u>-1 2 3</u> 4 5 6 7 8- {9 10 11 12} <em>13 14</em>

physics

<u>-1 2 3 </u>4 5 6 7 8-

biology

<u>-1 2 3 -</u><em>13 14</em> 15 <em>16 17 18 19 20</em>

there are 14 students in chemistry and 11 in biology, 5 of them are in both classes so 14-5= 9 and 11-5= 6 9+6=16. BUT 8 of those sixteen are also in -physics- so we arent counting them, we put those students in dashes. so lets count the students out of the dashes. so<em> twelve</em> of them are in chemistry and biology but not physics.

7) so how many are in just biology but not the other classes? lets look at the graph.

<u>1 2 3 </u>4 5 6 7 8 {9 10 11 12} <em>13 14</em> 15 <em>16 17 18 19 20</em>

chemistry

<u>1 2 3</u> 4 5 6 7 8 {9 10 11 12} <em>13 14</em>

physics

<u>1 2 3 </u>4 5 6 7 8

biology

<u>1 2 3 </u><em>13 14</em> 15 <em>16 17 18 19 20</em>

in the class of biology students 1 2 3 13 and 14 take other classes that leaves students 15 16 17 18 19 and 20. so 6 students take ONLY biology.

sorry its long but i hope this helps!

3 0
3 years ago
Steven works at the Pi Day food truck and he gets paid $9 per hour x. If 1 point
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P(x) =9x +30

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Segment HI has a length of 25.3 inches. It can be mapped to segment GO by a reflection over the x-axis followed by a reflection
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25.3, reflections do not change the length of something and if it can be mapped on to it they should be the same
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