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BlackZzzverrR [31]
4 years ago
10

Use the graph to determine the input value for which f(x)=g(x) is true.

Mathematics
1 answer:
Lostsunrise [7]4 years ago
3 0

Answer:

x = 1.5

Step-by-step explanation:

We just have to find the x-coordinate of the point of intersection between the two functions. From the graph, they intersect at (1.5, 2) so the answer is 1.5.

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A single sheet of 20-pound bond paper is about a tenth of a millimeter thick. Each time you fold an ordinary sheet of paper in h
Likurg_2 [28]

The paper would be 112589991 km thick if folded 50 times.

Step-by-step explanation:

Step 1; The sheet is \frac{1}{10}mm thick which equals 0.1 mm. Every time it is folded it doubles in thickness. So if folded once it becomes 0.2 mm thick, if folded thrice it becomes 0.4mm thick.

Thickness if folded once = 2^{1} × 0.1 mm

thickness if folded twice = 2^{2}× 0.1 mm

Thickness if folded thrice = 2^{3} × 0.1 mm.

So if the paper is folded nth fold it would be 2^{n} × 0.1 mm thick.

Step 2; Using the above formula, if the paper is folded 50 times the thickness would be equal to 2^{50} × 0.1 mm = 112,589,990,684,262.4‬ mm.

1 km is equivalent to 10^{6}mm so 112,589,990,684,262.4‬‬ mm = 112,589,990.6842624‬‬ km which is rounded off to 112589991 km.

6 0
4 years ago
Sam is flying a kite. The length of the kite string is 80 meters, and it makes an angle of 75° with the ground. The height of th
olya-2409 [2.1K]

Answer:

THE HEIGHT OF THE KITE FROM THE GROUND=80mtr

Step-by-step explanation:

let H be the height of kite from ground

\sin\theta=\frac{height}{hypotenuse}

\sin\90=\frac{H}{80}

1=\frac{H}{80}

H=80meters

5 0
4 years ago
I'm trying to do this problem, but I have no idea how to even start it:
zhannawk [14.2K]

Answer given in the picture above.

3 0
4 years ago
Find the area of this circle. Use
makvit [3.9K]

Answer:

78.5

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Use the identity below to complete the tasks:
quester [9]

Answer:

a=[2q^{2}r]

b=[3s^{2}t]

Step-by-step explanation:

we have

8q^{6}r^{3}+27s^{6}t^{3}

we know that

8q^{6}r^{3}=(2^{3})(q^{2})^{3}r^{3}=[2q^{2}r]^{3}

27s^{6}t^{3}=(3^{3})(s^{2})^{3}t^{3}=[3s^{2}t]^{3}

therefore

a=[2q^{2}r]

b=[3s^{2}t]

substitute

a^{3} +b^{3}=(a+b)(a^{2} -ab+b^{2})

[2q^{2}r]^{3} +[3s^{2}t]^{3}=([2q^{2}r]+[3s^{2}t])([2q^{2}r]^{2} -[2q^{2}r][3s^{2}t]+[3s^{2}t]^{2})

[2q^{2}r]^{3} +[3s^{2}t]^{3}=([2q^{2}r]+[3s^{2}t])([4q^{4}r^{2}] -6[q^{2}r][s^{2}t]+[9s^{4}t^{2}])

7 0
3 years ago
Read 2 more answers
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