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jekas [21]
3 years ago
7

Help with question 13

Mathematics
1 answer:
notsponge [240]3 years ago
4 0
Find the longest possible length of each piece of ribbon
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These are two similar triangles.<br> What does x equal?
tangare [24]

Answer:

112

Step-by-step explanation:

So, we know that a triangle's sides must add up to have a sum of 180. If you add up your given lengths and you should get 47, not including x. Then if you look at the left side/the one without a given length and it should be identical to the 21, so you know that side is 21. So now add 21 to 47, you should get 68. Finally, you subtract 68 from 180 like this: 180-68, and it should equal 112. So you're answer is 112.

3 0
2 years ago
Someone help me with this plss​
ladessa [460]
2.57 + 5x = 25.92 is the equation and when you solve for x you would get x = 4.67
3 0
3 years ago
A toy company makes rectangular sandboxes that measure 6 feet by 5 feet by 1.2 feet. A customer buys a sandbox and 40 cubic feet
Iteru [2.4K]

Answer:

Customer has bought too much sand

Step-by-step explanation:

Given that the dimensions (Length, Width and Height) of rectangular box are = 6,5, 1.2.

Volume of a rectangular box = Length * Width * Height

=> 6*5*1.2

=> 36 cubic feet

As mentioned, customer bought a sand box and <u>40 cubic feet of sand.</u>

So,

Volume of rectangular box (R) = 36 cubic feet

Volume of Sand (S) = 40 cubic feet

From analysis

S > R

Which shows that the capacity of sandbox is 36 cubic feet and volume is 40 feet that is a little greater than desired capacity.

Therefore, customer has bought too much sand.

4 0
3 years ago
Can someone please help me​
astra-53 [7]

Answer:

square 20 has 44 green squares

square 21 has 45 green squares

Step-by-step explanation:

To solve the problem, we need to observe the cases, and determine/define a rule for each case (odd number of sides, or even number of sides).

For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.

The side length is 3 for square 1, 4 for square 2, and so on.

Let

n= square number (1, 2,3...)

L = side length (3,4,5...)

G1(n) = function that gives the number of green squares for square n, n=odd

G2(n) = function that gives the number of green squares for square n, n=even

side length, L=n+2   ................(1)

G1(n) = twice the side length less one, as discussed above

G1(n) = 2L-1       now substitute L=n+2

G1(n) = 2(n+2) -1    simplify

G1(n) = 2n + 3

Check:

for n=1, square 1 has 2*1+3 = 5 green squares ... checks

for n=3, square 3 has 2*3+3 = 9... checks

for n=5, square 5 has 2*5+3 = 13 ....checks

For even squares, it is even easier, because

G2(n) = 2L = 2(n+2)

check:

for n=2, square 2 has 2(2+2) = 8 green squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

Fincally, we apply our formula to n=20 and n=21

square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars

square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares

5 0
3 years ago
39 ft/min= ____ yd/s explain your reasoning
dimulka [17.4K]

1yd=3ft...so 39ft=(39/3)=13yds

1mins =60seconds

13/60=0.22yds/secs

7 0
3 years ago
Read 2 more answers
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