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hoa [83]
2 years ago
13

Solve,y=5 and 3x +2y=-5

Mathematics
2 answers:
ddd [48]2 years ago
7 0

Answer:

x = -5

y = 5

Step-by-step explanation:

We are given the system of equations :

3x + 2y = -5

y = 5

This can be solved through the process of substitution, mainly because y in the second equation already has a value in of its own and can be replaced with the y in the first equation. Therefore :

3x + 2y = -5

y = 5

3x + 2(5) = -5

3x + 10 = -5

Solve for x, first by subtracting ten by both sides :

3x = -15

Divide 3 by both sides to isolate x :

x = -5

Now that we have x, we have solved our system of equations, but let's prove it first.

x = -5

y = 5

3(-5) + 2(5) = -5

-15 + 10 = -5

Therefore, our two variables are x = -5 and y = 5

Allushta [10]2 years ago
5 0

Answer:

x = -5

Step-by-step explanation:

I) Substitute:

3x + 2y = -5

3x + 2(5) = -5

3x + 10 = -5

II) Solve:

3x + 10 = -5

3x = -15 (Subtract 10 from both sides)

x = -5 (Divide both sides by 3)

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4 + 6^2 + 9 ∙ 3 - 10
Korvikt [17]
4+6^2+9•3-10

PEMDAS
P=parentheses
E=exponent
m=multiplication
d=division
a=addition
s=subtraction

*no parentheses in this equation
*exponent = 6^2
6^2=36
4+36+9•3-10
*multiplication = 9•3
9•3=27
4+36+27-10
*addition=4+36+27
4+36+27=67
67-10
*subtraction=67-10
67=10=57

ANSWER=57







please give brainliest itd mean a lot<3
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A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
irga5000 [103]

Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

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The answer is a point
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andre [41]

The volume of the planter that measures 60 l, 60 w. 60 h (D) 216000 cm³

<u>Explanation:</u>

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Width, w = 60 cm

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Volume, v = ?

We know,

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Substitute the values in the formula.

v = 60 X 60 X 60\\\\v = 216,000 cm^3

Therefore, the volume of the planter is 216,000 cm³

8 0
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