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3 years ago

Could someone pkease help. i am not sure if i have the correct answer, thanks

Mathematics

1 answer:

Mama L [17]3 years ago

5 0

I believe you are correct keep it up

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5. Marie buys 2 pounds of turkey and 1 pounds of ham. She pays a total of $20,

NARA [144]

Answer:Let t = cost per pound of turkey

Let h = cost per pound of ham

4t + 2h = 30

t = h - 1.5

substitute h-1.5 in place of t in 1st equation

4(h-1.5) + 2h = 30

4h - 6 + 2h = 30

6h - 6 = 30

6h = 36

h = 6

t = h-1.5 = 4.5

1 pound of turkey costs $4.50

1 pound of ham costs $6.00

----------

cost of 1 pound each $10.50

Step-by-step explanation:

8 0

2 years ago

What is the value of x ? <br> A : 15<br> B : 30 <br> C: 45 <br> D : 60

jenyasd209 [6]

Answer:

Step-by-step explanation:

So that is a 90 degree angle and there is an angle of 30. 30-90 is 60. 60 divided by 4 is 15. So thus the answer is A,15.

7 0

3 years ago

Read 2 more answers

Find the area of the square when z=3/2 answer

BartSMP [9]

Answer:

Step-by-step explanation:

The formula of an area of a square with side z:

A = z²

We have z = 3/2. Substitute:

A = (3/2)² = (3/2)(3/2) = 9/4

4 0

3 years ago

There are 360 people in a movie theater. Of these people, 35% bought student tickets, and 45% bought adult tickets. The rest bou

mafiozo [28]

Answer:

20%

Step-by-step explanation:

35+45 = 80

100-80=20

6 0

3 years ago

Read 2 more answers

Use integration by parts to find the integrals in Exercise.<br> x^3 ln x dx.

34kurt

Answer:

$\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

Step-by-step explanation:

We have been given an indefinite integral $\int \:x^3\:ln\:x\:dx$ . We are asked to find the value of the integral using integration by parts.

$\int\: u\text{dv}=uv-\int\: v\text{du}$

Let $u=\text{ln}(x)$ , $v'=x^3$ .

Now, we will find du and v as shown below:

$\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))$

$\frac{du}{dx}=\frac{1}{x}$

$du=\frac{1}{x}dx$

$v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}$

Upon substituting our values in integration by parts formula, we will get:

$\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$

Therefore, our required integral would be $\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

5 0

3 years ago