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DochEvi [55]
4 years ago
5

Please Help me ASAP read the problem below

Mathematics
1 answer:
SashulF [63]4 years ago
3 0

PART 1: Find the surface area of each section.

The formula for finding the area of a triangle is A = 1/2(b)(h); where b = base length and h = height. And the formula for finding the area of a rectangle is A = (l)(w); where l = length and w = width.

I'll start with the gold triangles, which are both the same size so I only need to calculate once.

A = 1/2(b)(h)

A = 1/2(7)(8.6)

A = 1/2(60.2)

A = 30.1 cm²

Now, I'll find the area of the blue triangles.

A = 1/2(b)(h)

A = 1/2(12)(7)

A = 1/2(84)

A = 42 cm²

Now, I'll find the area of the rectangular base.

A = (l)(w)

A = (12)(7)

A = 84 cm²

PART 2: Find the amount of foil in each color needed to cover 1 ornament.

There are two gold triangles, so all we have to do is multiply the area of the gold triangle by 2.

(30.1)(2) = 60.2 cm²

There are two blue triangles, so all we have to do it multiply the area of the blue triangle by 2.

(42)(2) = 84 cm²

And Olga will need 84 cm² of black foil to cover the rectangular base.

PART 3: Find the total amount of foil needed to cover 1 ornament and then 100 ornaments.

To find the total amount of foil needed, just add the amounts found in the previous step.

60.2 + 84 + 84 = 228.2 cm²

Olga will need 228.2 cm² of foil to cover 1 ornament.

(228.2)(100) = 22820 cm²

Olga will need 22, 820 cm² of foil to cover 100 ornaments.

PART 4: Fill in the blanks to give your answer.

Surface area of black face(s) = 84 cm²

Surface area of blue face(s) = 84 cm²

Surface area of gold face(s) = 60.2 cm²

Total surface area of 1 ornament = 228.2 cm²

Total surface area of 100 ornaments = 22, 820 cm²

Hope this helps!

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Help me please!!!!!!!!!
Maslowich

Answer:

a. -16/3 or -5 1/3

b. 16/3 or 5 1/3

c. -6

d. -6

e. 6

Step-by-step explanation:

5 0
3 years ago
What can you tell about two integers when their quotient is positive? negative? or zero?
Archy [21]
If the quotient is positive, we know that the two integers are either both positive or both negative.

If the quotient is negative, we know that one integer is positive and the other is negative.

If the quotient is zero, then we are dividing 0 by some non-zero integer.
3 0
3 years ago
A line includes the points (0, –6) and (–2, –10). What is its equation in slope-intercept form?
sergiy2304 [10]

Answer:

y = 2x - 6

Step-by-step explanation:

The slope-intercept equation is y = mx + b

where m is the slope and b is the point on the y axis when x=0 which is called the y intercept.

The slope is the rise over run.

Rise is the difference in the y values.

Run is the difference in the x values.

Rise = difference between -6 and -10  which is 4 units

Run = difference between 0 and -2 which is 2 units.

The slope would then be 4 over 2 or 4/2 which can be reduced down to 2.

If you visualize the line on the graph, the line slopes upward from left to right.  So the slope must be a positive number.

Slope = 2

The y intercept is when x=0.  One of the coordinates (0, -6) gives you that answer.  When x=0, y = - 6.  This is b.

Your equation would be y = 2x - 6

7 0
3 years ago
Read 2 more answers
Please help! my friend needs help with his grade 11 math homework! Thank you
likoan [24]

Answer:

a) the first negative number will be -1

b) A, B and C

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Step-by-step explanation:

3 0
3 years ago
The sum of 8th term of an A.P is 160 while to sum of the 20th term is 880. Find the 43rd term​
marshall27 [118]
Let a and d be the first term and common difference, respectively. Then

the 8th term is a+7d
the 20th term is a+19d

The sum of the first 8 terms is
(a)+(a+d)+(a+2d)+...+(a+7d) = 8a+28d

The sum of the first 20 terms is
(a)+(a+d)+(a+2d)+...+(a+19d) = 20a+190d

So

8a + 28d = 160
20a+ 190d = 880

40a + 140d = 800
40a + 380d = 1760
240d = 960
d = 4

8a + 112 = 160
8a=48
a =
6

The first term is 6 and the common difference is 4.

The 43rd term is a+42d = 6+42(4) = 6+168 = 174

The sum of the first 12 terms is
(a)+(a+d)+(a+2d)+...+(a+11d) = 12a+66d = 12(6)+66(4) = 72+264 = 336
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3 years ago
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