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Svet_ta [14]
3 years ago
12

Ying Ying wants to buy some petunias.

Mathematics
1 answer:
Crazy boy [7]3 years ago
5 0
2 petunias leave $6.50 on her wallet, and 8 leave $5.00. This means that 6 petunias cost $1.50, and that 1 petunia costs 25 cents.

.25 goes into 3.50 (the amount of money she has left in her wallet after buying 14 petunias) 14 times. 14 + 14 =28.

Ying Ying can buy 28 petunias at most.
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PLZZZZZZ HEEEEEEEEELLLLLPPPPP<br> (will mark brainliest for correct answers)
alexdok [17]

Answer:

y=4

Step-by-step explanation:

3(9) +4y= 43

27+4y=43

4y=16

y=4

6 0
2 years ago
Write the equation of the line that passes through the given points<br> (0,2)) and (-9, -3)
Irina18 [472]

Step-by-step explanation:

Y= m x+ C

m (slope ) = 2 - (_3) / 0 _( - 9 )

= 5 / 9

( O, 2 ) Satisfying equation

2= 5/9 ( 0) + C

C= 2

So eq. is

y= 5/ 9 X + 2

7 0
3 years ago
Ebi, Jose, Derell, and Asami measured their heights. Ebi's height was 2.5 cm greater than Jose's height. Jose's height was 3.1 c
irga5000 [103]
Start with assigning each person with a variable to represent their height

Ebi: e
Jose: j
Derell: d
Asami: a

Ebi'd height was 2.5 cm greater than Jose's height

j + 2.5 = e

Jose's height was 3.1 cm greater than Derell's

d + 3.1 = j

Derell's height is 0.4 cm less than Asami's height

a - 0.4 = d

Ebi is 162.5 cm tall

e = 162.5

So, plug in 162.5 into any of the above equations were there is a variable of e

j + 2.5 = e

j + 2.5 = 162.5

Subtract 2.5 from both sides of the equation

j = 160 cm

Jose's height is 160 cm

Now, plug in 160 into any of the above equations where there is a j

d + 3.1 = j

d + 3.1 = 160

Subtract 3.1 from both sides of the equation 

d = 156.9 cm

Derell's height 156.9 cm

so, plug in 156.9 into any of the above equations where there is a d

a - 0.4 = d

a - 0.4 = 156.9

Add 0.4 on both sides of the equation

a = 157.3 cm

Asami's height is 157.3 cm



7 0
3 years ago
Read 2 more answers
write an expression for the new balance using as few terms as possible. A checking account has a balance of -$126.89. A customer
Nutka1998 [239]

Answer:

New balance = -151.89 + 4.5x

(where x is the value of the smaller deposit)

Step-by-step explanation:

The inicial balance was -126.89.

Then, it were made two deposits, one with the value of x, and one that is 3 1/2 times the other, that is, 3.5x.

Then, the customer made a withdraw of $25.

So, The new balance can be calculated as:

New balance = inicial balance + x + 3.5x - 25

New balance = -126.89 + 4.5x - 25

New balance = -151.89 + 4.5x

8 0
3 years ago
First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-5U

q'=U-5

and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
3 years ago
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