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3 years ago

12

Use the distributive property to rewrite the expression –5(3x – 2).

1 answer:

Kazeer [188]3 years ago

8 0

The answer is -15x-10

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At the farmer's market,

Flura [38]

Answer:

Cantaloupe = $6, peach=$5

Step-by-step explanation:

$8x+3y=63\\2x+3y=27\\$

Subtract the first one from the second one

$8x-2x+3y-3y=63-27\\6x=36\\x=6\\$

Plug in:

$2(6)+3y=27\\12+3y=27\\3y=15\\y=5$

Hope this helps plz hit the crown :D

3 0

3 years ago

24 students and Jamall's class are wearing tennis shoes there are 30 students in his class tomorrow says that 70% of his class i

alexdok [17]

10% = 3
70% = 3 * 7
= 21 students

Jamall is wrong because 70% of the class = to 21 students.

5 0

3 years ago

A bakery sells 5 muffins for every 6 cookies it sells complete the table

tekilochka [14]

Answer:

5 muffin 6 cookie, 10 muffin 12 cookie, 15 muffin 18 cookie, 20 muffin 30 cookie

Step-by-step explanation:

think of it as a fraction like 5/6 is the same as 10/12 and 15/18 and so on

7 0

3 years ago

There are nine empty seats in a theater, and six customers need to find places to sit. How many different ways can these six sea

dybincka [34]

Answer: 60480

Step-by-step explanation:

Given : The number of empty seats in a theater = 9

The number of customers need to find places to sit =6

Since order matters here, so we use permutations

The permutations of n things taking r at a time is given by :-

$^nP_r=\dfrac{n!}{(n-r)!}$

Then, the number of ways to arrange 6 seat in 9 seats :-

$^9P_6=\dfrac{9!}{(9-6)!}=\dfrac{9\times8\times7\times6\times5\times4\times3!}{3!}\\\\=60480$

Hence, the number of ways to arrange 6 seat in 9 seats = 60480

3 0

3 years ago

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) <em>Large and small drains are opened</em>

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) <em>Only the small drain is opened</em>

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) <em>Only the big drain is opened</em>

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$

$t_{s} = 5.303\,h$

The draining time when only the small drain is opened is 5.303 hours.

7 0

3 years ago