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Mazyrski [523]
2 years ago
12

Helppppp me plzzzzzzzzzzz somebody

Mathematics
2 answers:
worty [1.4K]2 years ago
6 0
The answer above is correct my dude
Inessa05 [86]2 years ago
5 0

Answer:

(2,0)

Step-by-step explanation:

On the x-axis, the point is at 2. The point has a y-value of 0, as it's on the x-axis.

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What is 3/20 written as a percent?
alex41 [277]
Express it as a decimal. 

3/20 = 0.15 by division 

then move the decimal point twice to the right and affix the percent symbol 

thus, 15%.
6 0
3 years ago
Read 2 more answers
If f and g are continuous functions with f(0) = 2 and the following limit, find g(0). lim_(x->0)[2 f(x) - g(x)] = -3
ladessa [460]

We have by distributivity of limits over sums, and continuity of f(x) and g(x), that

\displaystyle \lim_{x\to0} \bigg(2f(x) - g(x)\bigg) = 2 \lim_{x\to0} f(x) - \lim_{x\to0} g(x) = 2 f(0) - g(0)

Given f(0)=2, it follows that

2^2 - g(0) = -3 \implies g(0) = \boxed{7}

5 0
1 year ago
What adds to 8 but multiplies to -100?
Pavlova-9 [17]
Factors of -100
-1,100 or 1,-100
-2,50 or 2,-50
-4,25 or 4,-25
-5,20 or 5,-20
-10,10
now through trial and error we add these together until we get 8 which is...
none of these, are you sure that you have the correct numbers?
5 0
3 years ago
Read 2 more answers
Ill give brainliest - please help ASAP<br><br> WITH THE WORK THO PLEASE?
ohaa [14]

Answer:

5.

we have.

11x=½(16x+16+50)

22x=16x+66

22x-16x=66

6x=66

x=66/6

x=11

6.

118°=½(6x-11+181)

236=6x+170

6x=236-170

x=66/6

x=11°

7.

34°=½(arcAK-arc LN)

34×2=(110-arc LN)

arc LN =110-68

arc LN=42°

8.

<L=½(arc EN-arc KM)

<L=½(139°-73°)

<L=33

4 0
3 years ago
If
Leno4ka [110]

Answer:

\frac{s^2-25}{(s^2+25)^2}

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given:\mathcal{L}[t \cos 5t]=(-1)F'(s) with F(s)=\mathcal{L}[\cos 5t].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s).

Solving for F(s) on the last equation, F(s)=\frac{s}{s^2+25}, then the Laplace transform we were searching is -F'(s)=\frac{s^2-25}{(s^2+25)^2}

3 0
3 years ago
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