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Tresset [83]
3 years ago
14

Town B is 40 km due north of town a what is the bearing of a from B ​

Mathematics
1 answer:
Dafna11 [192]3 years ago
3 0

Answer:

180°

Step-by-step explanation:

In bearing the protractor is placed in the North-South direction(eastside) thus directly north is on a bearing of 0°.After you mark the point B. A will be directly south which is on a bearing of 180°

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Masja [62]

Answer:

blue increased is addition

Step-by-step explanation:

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2 years ago
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A triangular prism has a base that is 6 cm by 4 cm and a height of 8 cm. If all dimensions are tripled, what happens to the volu
Montano1993 [528]

Answer:

The last option:  27 times the original volume.

Step-by-step explanation:

The dimensions are tripled so and  we are talking volumes so:

The factor of enlargement is 3^3 = 27.

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2 years ago
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Find the first term and the common ratio <br>third term =6 and seventh term= 96​
mamaluj [8]

Answer:

First term a₁ = 3/2 and common ratio r = 2

Step-by-step explanation:

We need to find the first term and common ratio while we are given third term = 6 and seventh term = 96

Since common ratio is required so, the sequence is geometric sequence

The formula used is: a_n= a_1r^{n-1}

We are given: third term = 6 i,e

a_3=a_1r(3-1)\\6=a_1r^2----eq(1)

Seventh term = 96

a_7=a_1r(7-1)\\96=a_1r^6----eq(2)

Dividing eq(2) and eq(3)

\frac{96}{6}=\frac{a_1r^6}{a1r2}\\16=\frac{r^6}{r^2}\\16=r^{6-2}\\=> r^4=16\\Taking \ fourth \ root\\ r=2,-2,2i,-2i\\ \ We \ will \ consider \ only \ positive \ value \ of \ r \ i.e \ \textbf{ r= 2}

So, Common Ratio r = 2

Finding First term using eq(1)

a_3=a_1r^2\\6=a_1(2)^2\\6=a_1(4)\\a_1= \frac{6}{4}\\a_1= \frac{3}{2}

So, First term a₁ = 3/2 and common ratio r = 2

6 0
3 years ago
Solve.<br> –51/2 + 63/4 + (-41/4)
VladimirAG [237]

Answer:

-20

Step-by-step explanation:

cause i did the math hope this helped :D

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A class at Middlebury school collected date on the types of movies students prefer. Complete each statement using the table.
likoan [24]

Based on the information of the table, you have:

1. The ratio is 105/150. By simplifying you get for the ratio 7/10.

2. The students that prefer action movies are 75+90 = 165 and the total numbe of students is 180+240 = 420. Then, the fraction of students who prefer action movies is:

165/420 = 11/28

3. The fraction of seventh graders students that prefer action movies is:

75/180 = 5/12

4. The percent of student that prefer comedy is:

105 + 150 = 255 total student that prefer comedy

420 total number of students

the fraction is:

(x/100)420 = 255

solve for x:

x = 255(100/420)

x = 60.71

the percent of students is 60.71%

5. The percent of eighth graders student who prefer action moveis is:

(x/100)240 = 90

x = 90(100/240)

x = 37.5

the percent of students is 37.5%

6. To determine which from the given grades has the greatest percent of student that prefer action movies, calculate the percent of student in seventh-grade:

(x/100)180 = 75

x = 75(100/180)

x = 41.66

the percent of student is 41.66%

then, seventh grade has the greatest percent of student that prefer action movies.

4 0
9 months ago
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