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3 years ago

Help on #26 please. State the domain and range

Mathematics

1 answer:

ladessa [460]3 years ago

4 0

Wouldnt the domain be x with the equal cross thingy 3

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How much larger is 6 times 10 to the 5th power compared to 2 times 10 to the 3rd power

lora16 [44]

Answer:

Either 10 times or 598,000.

Step-by-step explanation:

6 x 10 ^ 5 = 600,000

2 x 10 ^ 3 = 2,000

If we are figuring out the exact number, 600,000 - 2,000. If we are finding out how many powers larger, count.

600,000 - 2,000 = 598,000

600,000 is 10 times larger than 2,000.

See?

600,000

2,000

6 0

3 years ago

Write an equivalent expression to

mel-nik [20]

1. Distributive Property.
4(3a + 7) + 3(2a + 5)
12a + 28 + 6a + 20

2. Combine like terms
12a + 28 + 6a + 20
18a +48

The equivalent expression is 18a + 48. If a = anything then the expressions will still be equivalent.

5 0

3 years ago

How do you represent any number in an equation?

RideAnS [48]

Any unknown number in an equation you represent with a variable

5 0

3 years ago

Read 2 more answers

The following numbers are gas prices from 12 local gas stations. If Pilot builds another local gas station, what would the gas p

TEA [102]

Answer:

1.24

Step-by-step explanation:

4 0

3 years ago

How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title

kotegsom [21]

Let $t=\sqrt y$ , so that $t^2=y$ , $t^4=y^2$ , and $\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}$ . Then

$\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}$

Now let $y=\sqrt2\tan z$ , so that $\mathrm dy=\sqrt2\sec^2z\,\mathrm dz$ . Then

$\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C$

Transform back to $y$ to get

$\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C$

and again to get back a result in terms of $t$ .

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$

3 0

3 years ago