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BartSMP [9]
3 years ago
7

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

t to x from the second equation is substituted into the first equation. 3x + 2y = −21 x − 3y = 4
Mathematics
1 answer:
Svetllana [295]3 years ago
3 0
Hello,

\left \{ {{3x+2y=-21} \atop {x-3y=4}} \right. \\\\
 \left \{ {{3x+2y=-21} \atop {x=4+3y}} \right. \\\\
 \left \{ {{3(4+3y)+2y=-21} \atop {x=4+3y}} \right. \\\\

 \left \{ {{12+9y+2y=-21} \atop {x=4+3y}} \right. \\\\
 \left \{ {{12+11y=-21} \atop {x=4+3y}} \right. \\\\
 \left \{ {{11y=-21-12} \atop {x=4+3y}} \right. \\\\
 \left \{ {{11y=-33} \atop {x=4+3y}} \right. \\\\
 \left \{ {{y=-3} \atop {x=4+3(-3)}} \right. \\\\
 \left \{ {{y=-3} \atop {x=5}} \right. \\\\
Sol=\{(5,-3)\}\\\\




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The sum of liz's and melinda's ages is 28. if liz was 1/3 melinda's age 6 years ago, how old will liz be next year?
Mrrafil [7]
The first thing we must do for this case is to define variables:
 We have then:
 x: liz's age
 y: melinda's age
 We now write the system of equations:
 x + y = 28

x-6 = (1/3) (y-6)
 By solving the system graphically we have that the age of each one is currently:
 x = 10

 y = 18
 Next year, the age of liz is:
 x + 1 = 10 + 1 = 11

 Answer:
 
Next year, Liz's age is 11 years.
 
Note: see attached image for the graphic solution.

3 0
3 years ago
Consider the line =+3x6y−8 . what is the slope of a line parallel to this line? what is the slope of a line perpendicular to thi
Reil [10]
<span>Two lines are parallel if their slope is the same.
You want to write 8x + 4y = 5 in the form y = mx + b, where m represents the slope and b is the y-intercept.
We need to isolate y in the given equation. The number next to x is the slope.
8x + 4y = 5
4y = -8x + 5

y = (-8x +5)/4
y = -2x + 5/4
The slope of the line we want is -2.
Two lines are perpendicular if the slope of the first line times the slope of the second line produces a product of negative one.
Since our slope is -2, we know that -2 times 1/2 yields -1.
The slope of the line perpendicular is 1/2. </span>
7 0
3 years ago
Find the missing side of the triangle. Round your answers to the nearest tenth if necessary.
Alik [6]

Given:

In a right angle triangle, the measures of two legs are 5 ft and 5.6 ft. The measure of the hypotenuse is x.

To find:

The value of x.

Solution:

Pythagoras theorem: In a right angle triangle,

(Hypotenuse)^2=(Base)^2+(Perpendicular)^2

Using Pythagoras theorem, we get

(x)^2=(5)^2+(5.6)^2

x^2=25+31.36

x^2=56.36

Taking square root on both sides, we get

x=\sqrt{56.36}

x=7.50733

x\approx 7.5

The measure of missing side is 7.5 ft. Therefore, the correct option is D.

4 0
3 years ago
Ana drinks chocolate milk out of glasses that each hold 1/8 of a leader. she has 7/10 of a liter of chocolate milk in her refrig
PIT_PIT [208]

Amount of chocolate milk Ana drinks = \frac{1}{8} liter

Total amount of chocolate milk in refrigerator = \frac{7}{10} liter

We have to determine the total number of glasses she will pour \frac{1}{8} liter of milk from  \frac{7}{10} liter of milk.

Total number of glasses will be determined by dividing the total amount of milk by the amount of milk required for one glass

Total number of glasses = \frac{7}{10} \div \frac{1}{8}

= \frac{7}{10} \times 8

= \frac{56}{10}

= 5.6 glasses

So, 5.6 glasses will be poured from the given amount of chocolate milk.

3 0
3 years ago
The curve y = |x|/ 5 − x2 is called a bullet-nose curve. find an equation of the tangent line to this curve at the point (2, 2).
diamong [38]
We have the following curve:

y=\frac{\left | x \right |}{5-x^2}

So we need to find <span>an equation of the tangent line to this curve at the point (2,2). So let's find out if this point, in fact, belongs to the curve:

</span>If \ x=2 \\ \\ y=\frac{\left | 2 \right |}{5-2^2}=\frac{\left | 2 \right |}{5-4}=2. 
<span>
We also know that:

</span>\left | x \right |= \left \{ {{x \ if \ x \ \geq 0} \atop {-x \ if \ x\ \textless \ 0}} \right.<span>

Given that the point is:

</span>(2,2)\ that \ is: \\ x=2\ \textgreater \ 0

Then we will say that:

\left | x \right |=x

Therefore:

y=\frac{x}{5-x^2}

Computing the derivative:

y=\frac{x}{5-x^2} \\  \frac{dy}{dx}= \frac{(1)(5-x^2)-(-2x)x}{(5-x^2)^2}=\frac{5+x^2}{(5-x^2)^2}

So the derivative solved for x=2 is in fact the slope of the line at the point (2,2), then:

\frac{dy}{dx} |_{x=2}=\frac{5+x^2}{(5-x^2)^2}|_{x=2}=\frac{5+2^2}{(5-2^2)^2}=9 \\ \\ \therefore m=\frac{dy}{dx} |_{x=2}=9

Finally, the tangent line is:

y-y_1=m(x-x_1) \\ \therefore y-2=9(x-2) \\ \therefore \boxed{y=9x-16}

<em>This is shown in the figure below.</em><span>
</span>

8 0
3 years ago
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