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sergiy2304 [10]
3 years ago
8

Write 2^3 ÷ √ 2 as a single power of 2.

Mathematics
1 answer:
USPshnik [31]3 years ago
8 0

Answer:

Step-by-step explanation:

Numerator:  2^3

Denominator:  2^(1/2)

Combining these, we subtract 1/2 from 3, obtaining 2^(5/2)

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Answer: $6.50 + $2x ≤ $15.00

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1,654,990 round the 4 place value
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1,655,000 maybe I not entirely sure sorry
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H(x) = x3 + 7<br> Determine whether the function is one-to-one.
zubka84 [21]

Answer:

we conclude that the function is one-to-one.

Step-by-step explanation:

A function will a one-to-one function if it

  • passes the vertical line test to make sure it is indeed a function, and
  • also a horizontal line test to make sure it is it one-to-one.

In other words,

The function will be one-to-one if it passes the vertical line test, and also if the horizontal line only cuts the graph of the function in one place.

The reason is that there must be only one x-value for each y-value.

Given the function

h\left(x\right)\:=\:x^3\:+\:7

Have a look at the attached graph.

  • The red portion represents the graph of the function h\left(x\right)\:=\:x^3\:+\:7.
  • The green portion represents the graph of x=2 which is basically a vertical line test. Vertical line indicates that it cuts the cuts the graph of the function in one place. So it is clear that h\left(x\right)\:=\:x^3\:+\:7 is indeed a function.
  • The blue line represents the graph of y=9, which is basically a horizontal line test. Horizontal line indicates that it cuts the cuts the graph of the function in one place. So it is clear that h\left(x\right)\:=\:x^3\:+\:7 is a one-to-one function, as there is only one x-value for each y-value.

Therefore, we conclude that the function is one-to-one.

6 0
3 years ago
Farmer Jones, and his wife, Dr. Jones, decide to build a fence in their field, to keep the sheep safe. Since Dr. Jones is a math
alex41 [277]

Answer:\frac{8}{3}\times \sqrt{\frac{2}{5}}

Step-by-step explanation:

Given two upward facing parabolas  with equations

y=6x^2 & y=x^2+2

The two intersect at

6x^2=x^2+2

5x^2=2

x^2=\frac{2}{5}

x=\pm \sqrt{\frac{2}{5}}

area  enclosed by them is given by

A=\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx

A=\int_{\sqrt{-\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left ( 2-5x^2\right )dx

A=4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]

A=\frac{8}{3}\times \sqrt{\frac{2}{5}}

7 0
3 years ago
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