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3 years ago

I just need sum help:( and a tutor

Mathematics

1 answer:

seraphim [82]3 years ago

6 0

Answer:

The forth answer.

Try also transformation of curves bite (size) page.

I think it would reflect y less than x as we start x^2 for f(x) and usually 1 or 2 is the answer in quadratics. Once we have found y we can work out x

Step-by-step explanation:

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Karo-lina-s [1.5K]

Answer:

1. 74 because 8(10) - 6

2.4(4) - 5(3) = 1

3. 7(3)+8(4) * 2 = 106

4. A

5.a

6. 20.5

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8.a

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Step-by-step explanation:

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3 years ago

10!•8!/8!•7! simplify This equation

saveliy_v [14]

The answer is 4480 hope it helps

7 0

3 years ago

The tables show linear functions representing the estimated time it takes for the math and language arts portions of standardize

ozzi

Answer: D) The language arts function has the greater slope, which shows that the estimated time per question for language arts is greater than the estimated time per question for math.

Step-by-step explanation:

In the table shown, the estimated time for language arts is greater than the estimated time for math. They can answer 25 math questions in 87 minutes, whereas for language arts, they take 88 minutes to answer 20 questions. Therefore, the slope for LA is greater than the slope for Mathematics. Hope this helps :)

7 0

3 years ago

What is the volume of the figure?

prohojiy [21]

B
volume= length x width x height

6 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