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Anni [7]
4 years ago
10

A real world problem that is represented by the division problem 38 divided by 5 r3 in which it makes sense to round up the quot

ient up to 8
Mathematics
1 answer:
navik [9.2K]4 years ago
4 0
You have to get 38 water bottles and a number of water bottles in each pack are 5, the cost is represented by C.
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What ordered pair is a solution to y = x + 5 and x - 5y = -9?<br> I also need to show the work.
Margarita [4]

Answer:

(- 4, 1 )

Step-by-step explanation:

Given the 2 equations

y = x + 5 → (1)

x - 5y = - 9 → (2)

Substitute y = x + 5 into (2)

x - 5(x + 5) = - 9 ← distribute and simplify left side

x - 5x - 25 = - 9

- 4x - 25 = - 9 ( add 25 to both sides )

- 4x = 16 ( divide both sides by - 4 )

x = - 4

Substitute x = - 4 into either of the 2 equations and evaluate for y

Substituting into (1)

y = - 4 + 5 = 1

Solution is (- 4, 1 )

3 0
3 years ago
A child gets 20 heads out of 30 tosses of a coin. If he declared the chance of getting a head with that coin were 2/3, that woul
Komok [63]

Answer: Experimental probability

Step-by-step explanation:

There are two kinds of probability: Theoretical probability and Experimental probability.

To calculate theoretical probability we divide favorable outcomes by total outcomes.

To calculate experimental probability we divide number of times an event occurs by the total number of trials or times the activity is performed.

Here, A child gets 20 heads out of 30 tosses of a coin. If he declared the chance of getting a head with that coin were 2/3, which is dependent on the activity he performed, thus it is an experimental probability.

3 0
3 years ago
What are the zeros of x^3 - 18x^2 +107x -210
trapecia [35]

Answer:

(−7)(−6)(−5) so the zeros are 5, 6, &7

Step-by-step explanation:

4 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
Pleaseee help me with these i did some but just wanna make sure I'm in the same page if its right. please explain how you got it
Sergeeva-Olga [200]
1. P = 2(9) + 2(4.5)
P = 18 + 9
P = 27 m

2. P = 5.2(2) + 1.3(2)
P = 10.4 + 2.6
P = 13 ft

3. P = 12.9(2) + 4.7(2)
P = 25.8 + 9.4
P = 35.2 cm
5 0
3 years ago
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