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2 years ago

Lily is making a snack mix for her class.

Mathematics

1 answer:

oksano4ka [1.4K]2 years ago

4 0

Answer:

2,24,2,4

Step-by-step explanation:

1. to make 16 bags,we need to double all of our ingredients.

2. 6=2(3) so we need to multiply homy bags we can make with 2 cups of pretzels by 3

3. same as q1

4. we need to half the recipie which will half the outcome

hope it helps:)

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A square patio has an area of 122 square feet. How long is each side of the patio to the nearest 0.05?

jasenka [17]

Should be About 25 feet as your answer

3 0

3 years ago

16 2/3% of what number is 90?

zysi [14]

Answer:

540

Step-by-step explanation:

90 is 16 2/3% of 540

7 0

3 years ago

Jose has saved $75 to take tutoring classes. He must pay a registration fee of $15 and then an additional $6 for each class.

bonufazy [111]

Answer:

15+6x=75

Step-by-step explanation:

15 is the starting fee and it is 6$ per class so it would be 6x+15=75

4 0

3 years ago

Read 2 more answers

1. Three apples, a, and two peaches, p, sell for $1.25.

Simora [160]

Answer:

$0.25

Step-by-step explanation:

You can use systems of equations to solve this.

3a+2p = 1.25

5a+p = 1.5

Multiply the second equation by 2.

3a+2p = 1.25

10a+2p = 3

Subtract the bottom from the top

-7a= -1.75

Divide by -7

a= 0.25

Each apple is $0.25 (in this case, the peach also costs $0.25)

5 0

3 years ago

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