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kati45 [8]
3 years ago
7

This is my question it is in the screen shot please help me.

Mathematics
1 answer:
PolarNik [594]3 years ago
3 0
4 in = 10.2 cm
7kg = 15.4 lbs
6 gal = 22.7 L
66 ft = 19.8 m
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PLEASE HELP ME I WILL MARK BRAINLYEST
Korvikt [17]

                               Question 9:

Given the equations

5x+2y=6;

4x-8y=0

solving the system of the equations

\begin{bmatrix}5x+2y=6\\ 4x-8y=0\end{bmatrix}

\mathrm{Multiply\:}5x+2y=6\mathrm{\:by\:}4\:\mathrm{:}\:\quad \:20x+8y=24

\mathrm{Multiply\:}4x-8y=0\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:20x-40y=0

\begin{bmatrix}20x+8y=24\\ 20x-40y=0\end{bmatrix}

20x-40y=0

-

\underline{20x+8y=24}

-48y=-24

\begin{bmatrix}20x+8y=24\\ -48y=-24\end{bmatrix}

-48y=-24

\frac{-48y}{-48}=\frac{-24}{-48}

y=\frac{1}{2}

y = 0.5 ( 0.5 is the result of rounding 0.5 to the nearest 0.01)

\mathrm{For\:}20x+8y=24\mathrm{\:plug\:in\:}y=\frac{1}{2}

20x+8\cdot \frac{1}{2}=24

20x+4=24

\frac{20x}{20}=\frac{20}{20}

x=1.00 ( 1.00 is the result of rounding 1 to the nearest 0.01 )

So, the solution is:

(x, y) → (1.00, 0.5)

                            Question 10)

Similarly the question 10 can be solved.

Given the equations

5x+2y=9;\:4x-3y=44

solving

\begin{bmatrix}5x+2y=9\\ 4x-3y=44\end{bmatrix}

\mathrm{Multiply\:}5x+2y=9\mathrm{\:by\:}4\:\mathrm{:}\:\quad \:20x+8y=36

\mathrm{Multiply\:}4x-3y=44\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:20x-15y=220

\begin{bmatrix}20x+8y=36\\ 20x-15y=220\end{bmatrix}

20x-15y=220

-

\underline{20x+8y=36}

-23y=184

\begin{bmatrix}20x+8y=36\\ -23y=184\end{bmatrix}

-23y=184

y=-8.00   ( -8.00 Rounded to the nearest 0.01 or  the Hundredths Place )

\mathrm{For\:}20x+8y=36\mathrm{\:plug\:in\:}y=-8

20x+8\left(-8\right)=36

20x=100

x=5.00  ( 5.00 Rounded to the nearest 0.01 or  the Hundredths Place )

So, the solution is:

(x, y) → (5.00, -8.00)

8 0
3 years ago
X(x-4)+y(y-10)=-29 x;y=?
zaharov [31]
X sqrd-4x+y sqrd-10= -29
y sqrd-10y = -29-X sqrd + 4X
y(y-10) = -29-X sqrd +4X
y = - 29 -X sqrd +4X ÷( y- 10)
6 0
3 years ago
Can someone please help me with question 9
nataly862011 [7]

There are 24 oranges and 36 apples.

Each basket has to have the same number of fruit

So lets say that each basket will have exactly one orange and apple.

To get the maximum number of baskets, we have to use either all of the apples or all of the oranges.

You can't have 36 baskets, because you only have 24 oranges, even though there are enough apples.

But you can 24 baskets, because there are enough apples and oranges.

So you can use all of the oranges to make 24 baskets  (and you'll be left with 12 apples)

Answer: F. 24.

4 0
3 years ago
Help it’s my last question
Digiron [165]

Answer:

alr so basically what you want to do is multiply the numbers to get the answer

Step-by-step explanation:

6 0
3 years ago
Use long division to find the solution.
Kruka [31]
90 r 5 is the answer
3 0
3 years ago
Read 2 more answers
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