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3 years ago

WILL GIVE BRAINLIEST! what is the value of -x - y if -x=(-1/5) and y=(-7/15)

Mathematics

1 answer:

omeli [17]3 years ago

4 0

Answer: 20

Step-by-step explanation:

since i helped can i have brainlist please :D

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Why is adding 3/9 and 6/9 different from multpying the two fractions? Explain

tankabanditka [31]

<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ When you add fractions, you have to have the same denominator whereas, when you multiply them, you don't have to. (In this case the denominators are already the same.) Also, multiplying the fractions would make the result smaller whereas adding them would make it larger.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0

4 years ago

Please help it is a test

irakobra [83]

Answer:

c. -17

Step-by-step explanation:

We have the function f(x) = -2t^2 + 1 and we need to solve for x if f(-3).

Substitute the x in the function with the x in the equation :

f(-3) = -2(-3)^2 + 1

Solve for the answer :

f(-3) = -2(9) + 1

f(-3) = -18 + 1

f(-3) = -17

Therefore, c. -17 is the answer.

7 0

3 years ago

Find the consecutive integers even integers whose sum is 264

coldgirl [10]

Name the first even integer you are looking for $x$ . The next even number is $x+2$ .

Their sum is $264$ , so $x+x+2=264$ . That means $2x=262$ , hence $x=131$ .

So, the desired numbers are $131$ and $133$ .

5 0

4 years ago

What is the ratio 12/21 equivalent to

Amiraneli [1.4K]

4/7, because both the numerator and denominator can be divisible by 3.

7 0

3 years ago

Read 2 more answers

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