What is 1.15% of 26?

Megan has $45.12 in her purse. She spends $7.89 for lunch and $21.25 for a pair of sunglasses.

tekilochka [14]

Answer:

15.98 dollars

Step-by-step explanation:

its really easy

we need to subtract the total amount that she has spent from the total amount that she had in her purse

the total amount in Megan's purse=$45.12

total amount that Megan spent=$7.89+$21.25 =$29.14

now,

the exact amount that Megan has left = $45.12-$29.14=$15.98

5 0

2 years ago

What is 33+ 50<br> make as brainlest

Vika [28.1K]

Answer:

the answer is 83 because 5+3=8 and 3+0=3

8 0

3 years ago

Read 2 more answers

Please help on this one!! I need it in ASAP :(

lisov135 [29]

Answer:

Hey there! here's your answer

Step-by-step explanation:

Speed = distance/Time

Speed = 3/5 ÷ 1/2

Speed = 3/5 * 2

Speed = 6/5 mph or 1.2 mph.

5 0

3 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago

A real estate company wants to build a parking lot along the side of one of its buildings using 800 feet of fence. If the side a

Elden [556K]

Answer:

80,00 $ft^{2}$

Step-by-step explanation:

According to my research, the formula for the Area of a rectangle is the following,

$A = L*W$

Where

A is the Area
L is the length
W is the width

Since the building wall is acting as one side length of the rectangle. We are left with 1 length and 2 width sides. To maximize the Area of the parking lot we will need to equally divide the 800 ft of fencing between the <u>Length and Width.</u>

800 / 2 = 400ft

So We have 400 ft for the length and 400 ft for the width. Since the width has 2 sides we need to divide 60 by 2.

400/2 = 200 ft

Now we can calculate the maximum Area using the values above.

$A = 400ft*200ft$