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lana66690 [7]
2 years ago
13

What additional doubles fact can help you find 4+3? Explain how you know

Mathematics
1 answer:
Anna35 [415]2 years ago
5 0

Answer:

Either 4+4=8 then take 1 away to get 7 or 3+3=6 then add 1 to get 7.

Step-by-step explanation:

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in triangle pqr the measure of r=90 rp=3.7 and pq=9.5 find the measure of <Q to the nearest tenth of a degree​
Rasek [7]

Answer:22.9

Step-by-step explanation:

5 0
2 years ago
⅝-(-¾)= I don't know the answer plz help me
Leto [7]
A negative and a negative together make a positive so you are basically adding. so you have to find a common denominator which is 8 so you end up with 5/8 + 6/8 = 11/ 8 = 1 and  3/8
5 0
3 years ago
Read 2 more answers
45.50 is 30% of what number
m_a_m_a [10]

Answer:

31.85

Step-by-step explanation:

to subtract 30 percent from 45.50 just divide the percentage (30) by 100 subtract this value form 1 then multiply it by the initial value (45.50) so

final value = 45.50 x ( 1 - \frac{30}{100} )

final value = 45.50 x ( 1 - 0.3)

final value = 45.50 x (0.7)

final value = 31.85  

subtract 30% is the same as discount 30% this operation could be for example a 30% discount on 45.50

note: 0.3 can be found just by moving the decimal point 2 places to the left \frac{30}{100} = 0.3 as a decimal

3 0
2 years ago
The area of a trapezoid is 30 cm. If
Arisa [49]

The height of the trapezoid is 2 cm

7 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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