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MArishka [77]
3 years ago
13

OKAY, PEOPLE BIG NEWS AS I HAVE ALREADY MADE CLEAR I GOT EXAMS NEXT WEEK!!! I JUST GOT THE UPDATE AND ON MY MATH EXAM AND ITS 31

QUESTIONS FREAKING HARD AND ON THE ENGLISH 45 FREAKIN HARD QUESTIONS!!!! I DONT KNOW THE SPECIFICS ON THE BIOLOGY BUT I KNOW ITS GONNA BE BIG AND HARD!!!! I AM FAILING I REALLLLYYY NEED THISSS PASS RIGHT HERE SO, PEOPLE HERES HOW ITS GONNA GO: WHEN I POST A QUESTION I WILL NEED A CORRECT AND RIGHT ANWSER QUICKLY! I WILL NEED HELP PLEASEEEEEEEEEEEE!!!!!!!!!!!!!!!!!!!!!!!!.
I JUST WANNA PUT THAT OUT THERE FOR HEADS UP
Mathematics
2 answers:
bixtya [17]3 years ago
4 0

Answer: wow thats a lot of questions

Step-by-step explanation:

erica [24]3 years ago
3 0

Answer:

I'll try to do my best man, with all of my warnings

Step-by-step explanation:

You might be interested in
Select the name of the vector and complete its component form.
love history [14]

Answer:

  GH = (5, -3)

Step-by-step explanation:

The horizontal extent of the vector is 5 squares; the vertical extent is 3 squares. H has a lower y-value than G, so the vertical component is -3.

  GH = (5, -3)

8 0
3 years ago
9) Three times a number added to 12 gives -6. Find the number.​
KatRina [158]

Answer:

The number is -6.

Step-by-step explanation:

Variable x = a number

Set up an equation:

3x + 12 = -6

Isolate variable x:

3x = -18

x = -6

Check your work:

3(-6) + 12 = -6

-18 + 12 = -6

-6 = -6

Correct!

7 0
3 years ago
Read 2 more answers
NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.
Fudgin [204]

Answer:

The rock splashes down when h(t) = 0. Set h(t) = 0 and solve for t using quadratic formula.

0 = − 4.9t^2 + 178t + 325

t = 38.07, -1.74

(Throw away the negative answer, since time can't be negative.)

t = 38.07 sec

Please Mark Brainliest If This Helped!

4 0
2 years ago
Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len
Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> )   ==>   d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em>   ==>   d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt

=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt

=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt

=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt

=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}

5 0
3 years ago
Evaluate. 5x{(9 +2) X (6 + 3)} + 15​
Tju [1.3M]

Answer:15(3x11^+1)

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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