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kifflom [539]
3 years ago
15

I need help!! Please

Mathematics
1 answer:
Andre45 [30]3 years ago
8 0
I don’t know sorry mannnn
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A set of equations is given below:
Archy [21]
The answer is A because they are equal to eachother
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4 years ago
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What is the solution to the system of equations below?
polet [3.4K]

Answer:

X = 4 & Y = 3

Step-by-step explanation:

1. Multiply the first equation by -2 & Multiply the second equation by 1

-2 (2x+3y=17)

1  (3x+6y=30)  

Becomes:

−4x−6y=−34

3x+6y=30

Add these equations to eliminate y:

−x=−4

2. Then solve −x= −4 for x:

     −x=−4

    -x/−1 = −4/−1

(Divide both sides by -1)

x=4

Now that we've found x, plug it back in to solve for y.

Write down an original equation:

2x+3y=17

Substitute (4) for (x) in 2x+3y=17:

(2)(4)+3y=17

3y+8=17(Simplify both sides of the equation)

3y+8+−8=17+−8(Add -8 to both sides)

3y=9

3y/3 = 9/3

(Divide both sides by 3)

y=3

6 0
3 years ago
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John wants to gain weight for his wrestling team so he can compete in a different weight class. He currently weighs 187 pounds,
s344n2d4d5 [400]

Answer:

The equation would be 1.5w + 187 = y.

Step-by-step explanation:

1.5 lbs per week

already 187

The variable would be 'w'

so 1.5w +187 = y

6 0
3 years ago
The ordered pair below is a solution to which equation?
Helga [31]
Hello,

One solution is y=x/9
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6 0
3 years ago
The temperature fell from 0 Degrees Fahrenheit to 15 and one-half Degrees Fahrenheit below 0 in 5 and three-fourths hours. Wen t
Kryger [21]

Answer:

The correct answer will be:

-\dfrac{62}{23}

Step-by-step explanation:

It is given that :

Initial temperature, T_1 = 0^\circ F

Final temperature,

T_2 = -15\dfrac{1}{2}^\circ F\\\Rightarrow T_2 = -\dfrac{15\times 2+1}{2} ^\circ F\\\Rightarrow T_2 = -\dfrac{31}{2} ^\circ F

Time taken :

5\dfrac{3}{4}\ hrs = \dfrac{5 \times 4+3}{4}\ hrs = \dfrac{23}{4}\ hrs

Change in temperature per hour:

\dfrac{\text{Difference of temperature}}{\text{Total Time Taken}}\\\Rightarrow \dfrac{T_2-T_1}{\text{Total Time Taken}}

Putting the values of temperatures and time:

\dfrac{\dfrac{-31}{2}-0}{\dfrac{23}{4}}\\\Rightarrow \dfrac{\dfrac{-31}{2}}{\dfrac{23}{4}}\\\Rightarrow \dfrac{-31 \times 4}{2 \times 23}} \text{---- Error done by Wen at this step}\\\Rightarrow \dfrac{-31 \times 2}{23}}\\\Rightarrow \dfrac{-62}{23}}

The error done by Wen was during calculating the values of fraction.

So, the correct answer is :\frac{-62}{23}} instead of \frac{-713}{8}

5 0
4 years ago
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