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MrRissso [65]
3 years ago
13

Ratios for 42 and 54

Mathematics
2 answers:
KiRa [710]3 years ago
8 0

Answer:

The ratio is easy to find!

Ratios are:

42:54

42 to 54

42/53

Step-by-step explanation:

            May I have brainliest I'm trying to level up!

                        </3 PureBeauty

seropon [69]3 years ago
3 0

Answer:30 and 7

Step-by-step explanation:

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50 POINTS AND BRAINLIEST!!! NEED HELP NOW!!
Ad libitum [116K]

1. would have to be team E because add up all the numbers 1058/8 (which is how many numbers that is there) 132.25.

2. (More of an explanation if you still don't understand)

52+52+53+54+55+55=321/6=53.5

52+53+54+55+57+57=328/6=54.66 repeated rounded to 54.7 so your answer is 1.2

3 0
2 years ago
HELP ASAP!!!! 30 POINTS
Vedmedyk [2.9K]

Answer:

20.8 ft²

Step-by-step explanation:

area = (ab sin C)/2

area = (6 ft × 8 ft × sin 60°)/2

area = 20.8 ft²

6 0
2 years ago
Julia and Cody are working together on solving math problems. For every one problem that Julia
slega [8]

Answer:

Cody has solved (12 × 12) = 144 problems.

Step-by-step explanation:

For every one problem that Julia completes, Cody completes twelve.

If Julia Completes x problems and Cody completes y problems, then we can write y = 12x ........ (1)

Now, given that the number of problems solved by Cody is one hundred twenty more than two times the number of problems solved by Julia.

Hence, 2x + 120 = y ......... (2)

Now, from equations (1) and (2) we get,  

2x + 120 = 12x

⇒ 10x = 120

⇒ x = 12

Therefore, Cody has solved (12 × 12) = 144 problems. (Answer)

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
find all the solutions of sec theta - 2 =0. It says use "or" as necessary and gives an example of kpi.
swat32
Sec t (theta) - 2 = 0
sec t = 2
1/ cos t = 2
cos t =  1/2
Answer:
t = π / 3 + 2 k π ,  or :  t = 5 π / 3 + 2 kπ,
k ∈ Z
5 0
2 years ago
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