Which could be the resulting equation when elimination is used to solve the given system of equations ?

Answer:

45kg

Step-by-step explanation:

60.3/4=45

Answer:

(x+y)+2

hope this helps

Answer:

see below

Step-by-step explanation:

Slope intercept form is y = mx+b where m is the slope and b is the y intercept

y = x-7

m = 1

b = -7

y = 4

m=0

b = 4

y = -2x+3

m = -2

b = 3

Answer:

Δy = 5  Δx = 1

slope = [tex]\frac{5}{1}[/tex] = 5

Step-by-step explanation:

Answer:

what is it's perimeter?

Step-by-step explanation:

21/1

When a denominator is equal to 1, the fraction might be simplified, using this formula:

a/1=a

21/1=21

3/12

Reduce fraction

3/12=3÷3/12÷3

=1/4 or 0.25

Answer:

62

Step-by-step explanation:

12+12+3+3+7+10+15

you just add all the sides

Answer:

the answer is c you will thank us later

Answer:  Choice C)  [tex]\frac{8+2\sqrt{6x}}{8-3x}[/tex]

======================================

Work Shown:

[tex]y = \frac{4}{4-\sqrt{6x}}\\\\y = \frac{4(4+\sqrt{6x})}{(4-\sqrt{6x})(4+\sqrt{6x})}\\\\y = \frac{4(4+\sqrt{6x})}{4^2 - (\sqrt{6x})^2}\\\\y = \frac{4(4+\sqrt{6x})}{16-6x}\\\\y = \frac{2*2(4+\sqrt{6x})}{2(8-3x)}\\\\y = \frac{2(4+\sqrt{6x})}{8-3x}\\\\y = \frac{8+2\sqrt{6x}}{8-3x}\\\\[/tex]

This shows why choice C is the answer.

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Notes:

Answer:

56 months 12+12x2=56 hope this helps

Step-by-step explanation:

Answer:

The solution to the system is (-2,6)

Step-by-step explanation:

I said the solution to the system is (-2,6) because that is where both lines cross, meaning that's the solution.

Hope this is hepful.

Answer:

The cost of tuition as a function of x, the number of years since 1990, is C(x)= 14*(x-1990) + 95

Step-by-step explanation:

A linear function is a polynomial function of the first degree that has the following form:

y= m*x + b

where

So, in this case: C(x)= m*( x-1990) + b where x is the number of years since 1990.

Given the coordinates of two points, it is possible to determine the slope m of the line from them using the following formula:

[tex]m=\frac{y2 - y1}{x2 - x1}[/tex]

In this case, you know that in 1990, the cost of tuition at a large Midwestern university was $95 per credit hour. And in 1999, tuition had risen to $221 per credit hour. So:

So the value of m is:

[tex]m=\frac{221 - 95}{1999 - 1990}[/tex]

[tex]m=\frac{126}{9}[/tex]

m= 14

So C(x)= 14*( x-1990) + b. In 1999, tuition had risen to $221 per credit hour. Replacing:

221= 14*(1999 - 1990) + b

221= 14*9 +b

221= 126 + b

221 - 126= b

95= b

Finally, the cost of tuition as a function of x, the number of years since 1990, is C(x)= 14*(x-1990) + 95

Answer:

5x=13.5

Step-by-step explanation:

Well, we know that 1x=2.7. So, if we want 5x we need to multiply 2.7*5 becuase 1x=2.7, so 5x=2.7*5.

Multiply. 2.7*5=13.5

So, 5x=13.5.

Hope this helps!

Answer:

13.5

Step-by-step explanation:

5x = ?

Replace 'x' with 2.7.

5(2.7) = ?

5(2.7) = 13.5

Hope this helps.

Answer and Step-by-step explanation:

The answer is -7 (A.)

This is determined by looking at the graph. If you to to the x-point of -3 units, you will see there is a point at a y-value of -7 (units down).

#teamtrees #PAW (Plant And Water)

Answer:

No

Step-by-step explanation:

This relation is a function because the x values have a repeating number (15).

Answer:

Option (1)

Step-by-step explanation:

We have to graph the system of inequalities given.

y > 2x + 1 -----(1)

y > |x| --------(2)

For inequality (1),

Solution area will be the area above the dotted line y = 2x + 1

Similarly, solution area of the second inequality will be the area above the dotted lines of y = |x|

Solution area of the system of inequalities will be the common area of both the graphs of the given system.

That will lie in quadrants 1 and 2.

Option (1) will be the answer.

Answer:

79

Step-by-step explanation:

78 +2 ÷ 2

PEMDAS says divide first

78 + (1)

Then add

79

Answer:

y = (x + 1)²

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (- 1, 0) , then

y = a(x + 1)² + 0

To find a substitute (0, 1), the y- intercept into the equation

1 = a(0 + 1)² = a

y = (x + 1)²

Answer:

t < -3

Step-by-step explanation:

-3t + 11 > 20

subtract 11 from both sides

-3t > 9

Divide both sides by -3

when multipying or dividing by a negative the inequality must be reversed

t < -3

Step-by-step explanation:

answer is in photo above

Answer:

y = 3x - 30

Step-by-step explanation:

First, put both equations into slope-intercept form.

x + 3y - 4 = 0

3y = -x + 4

y = (-1/3)x + (4/3)

2x + 5y - 20 = 0

5y = -2x + 20

y = (-2/5)x + 4

For a line to be perpendicular to y = (-1/3)x + (4/3), its slope should be the opposite and reciprocal of this line's slope.

The slope of y = (-1/3)x + (4/3) is (-1/3). The slope of a perpendicular line is 3.

To find the x-intercept of y = (-2/5)x + 4, plug in 0 for y and find x.

0 = (-2/5)x + 4

x = 10

The equation of the line right now is y = 3x + b

Plug in (10, 0) to find the y-intercept.

0 = 3(10) + b

b = -30

The equation of the line is y = 3x - 30

Answer:

-2< x< 8

Step-by-step explanation:

|x-3|<5 <=> -5<x-3<5 <=> -2<x<8

..

Answer:

-2x<8

Step-by-step explanation:

|x-3|<5<=> -5<x-3<5 <=>-2<x<8

.....

Answer:

1 square unit

Step-by-step explanation:

Area of the shaded region

= Area of the triangle with base 3 units and height 1 unit - Area of triangle with base 1 unit and height 1 unit.

= 1/2 *3*1 - 1/2 *1*1

= 1.5 - 0.5

= 1 square unit

Answer:

44

Step-by-step explanation:

divide 24 by 6 then multiply it by all the number then add it all

please make me brainliest

Answer:

[tex]-2 < 2x - 3 < 1 \ = \ \frac{1}{2} < x < 2[/tex]

Step-by-step explanation:

[tex]-2 < 2x - 3 < 1\\\\-2 + 3 < 2x - 3 + 3 < 1 + 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ adding \ by\ 3 \ ]\\\\1 < 2x + 0 < 4\\\\1 < 2x < 4\\\\\frac{1}{2} < \frac{2x}{2} < \frac{4}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ divide \ by \ 2 \ ]\\\\\frac{1}{2} < x < 2[/tex]

Answer:

1/2 < x < 2

Step-by-step explanation:

-2<2x-3<1​

Add 3 to all sides

-2+3<2x-3+3<1​+3

1<2x<4

Divide all sides by 2

1/2 < 2x/2 <4/2

1/2 < x < 2

Answer:

Our two numbers are:

[tex]2+4\sqrt{2} \text{ and } 4\sqrt{2}-2[/tex]

Or, approximately 7.66 and 3.66.

Step-by-step explanation:

Let the two numbers be a and b.

One positive real number is four less than another. So, we can write that:

[tex]b=a-4[/tex]

The sum of the squares of the two numbers is 72. Therefore:

[tex]a^2+b^2=72[/tex]

Substitute:

[tex]a^2+(a-4)^2=72[/tex]

Solve for a. Expand:

[tex]a^2+(a^2-8a+16)=72[/tex]

Simplify:

[tex]2a^2-8a+16=72[/tex]

Divide both sides by two:

[tex]a^2-4a+8=36[/tex]

Subtract 36 from both sides:

[tex]a^2-4a-28=0[/tex]

The equation isn't factorable. So, we can use the quadratic formula:

[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

In this case, a = 1, b = -4, and c = -28. Substitute:

[tex]\displaystyle x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-28)}}{2(1)}[/tex]

Evaluate:

[tex]\displaystyle x=\frac{4\pm\sqrt{128}}{2}=\frac{4\pm8\sqrt{2}}{2}=2\pm4\sqrt{2}[/tex]

So, our two solutions are:

[tex]\displaystyle x_1=2+4\sqrt{2}\approx 7.66\text{ or } x_2=2-4\sqrt{2}\approx-3.66[/tex]

Since the two numbers are positive, we can ignore the second solution.

So, our first number is:

[tex]a=2+4\sqrt{2}[/tex]

And since the second number is four less, our second number is:

[tex]b=(2+4\sqrt{2})-4=4\sqrt{2}-2\approx 3.66[/tex]

Answer:

[tex]2+4\sqrt{2}\text{ and }4\sqrt{2}-2[/tex]

Step-by-step explanation:

Let the large number be [tex]x[/tex]. We can represent the smaller number with [tex]x-4[/tex]. Since their squares add up to 72, we have the following equation:

[tex]x^2+(x-4)^2=72[/tex]

Expand [tex](x-4)^2[/tex] using the property [tex](a-b)^2=a^2-2ab+b^2[/tex]:

[tex]x^2+x^2-2(4)(x)+16=72[/tex]

Combine like terms:

[tex]2x^2-8x+16=72[/tex]

Subtract 72 from both sides:

[tex]2x^2-8x-56=0[/tex]

Use the quadratic formula to find solutions for [tex]x[/tex]:

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex] for [tex]ax^2+bx+c[/tex]

In [tex]2x^2-8x-56[/tex], assign:

Solving, we get:

[tex]x=\frac{-(-8)\pm \sqrt{(-8)^2-4(2)(-56)}}{2(2)},\\x=\frac{8\pm 16\sqrt{2}}{4},\\\begin{cases}x=\frac{8+16\sqrt{2}}{4}, x=\boxed{2+4\sqrt{2}} \\x=\frac{8-16\sqrt{2}}{4}, x=\boxed{2-4\sqrt{2}}\end{cases}[/tex]

Since the question stipulates that [tex]x[/tex] is positive, we have [tex]x=\boxed{2+4\sqrt{2}}[/tex]. Therefore, the two numbers are [tex]2+4\sqrt{2}[/tex] and [tex]4\sqrt{2}-2[/tex].

Verify:

[tex](2+4\sqrt{2})^2+(4\sqrt{2}-2)^2=72\:\checkmark[/tex]

Answer:

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