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2 years ago

Solve each system by graphing 4x-y=9 and x-3y=16

Mathematics

2 answers:

Fynjy0 [20]2 years ago

5 0

\left[x \right] = \left[ 16+3\,y\right][x]=[16+3y] totally graphic doing ur sheets

expeople1 [14]2 years ago

5 0

Answer:

x = 1

y = -5

( 1 , -5 )

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Please help solve: Graph y-4=2(x-2)

xeze [42]

Answer:

Graph y = 2x

Step-by-step explanation:

First, let's get the equation into standard form. Distribute the 2 on the right.

$y-4=2x-4$

Next, we want the variable "y" to be alone, so we at 4 to both sides.

$y-4+(4)=2x-4+(4)\\\\y=2x$

That is our equation in standard y = mx + b form. "m" is our slope, while "b" is our y-intercept. Above , we don't have a value for b, therefore the line passes through the origin.

We do, however, have a slope, which can be thought of as $\frac{rise}{run}$ or rise over run. To represent this, we can rewrite our slope as:

$y=\frac{2}{1}x$

Meaning in each interval, the line goes up by 2 units, and moves forward by 1.

3 0

3 years ago

If three times a number minus 2 equals 13, what is the number?

Nana76 [90]

Change the messy words into numerals
3 times a number minus 2    equals    13
3 × n - 2 = 13

3n - 2 = 13
take 2 to the other side
3n - 2 + (2) = 13 + (2)
3n = 15

Divide by 3 on either sides to isolate n
$\frac{3n}{3}$ = $\frac{15}{3}$
3 and 3 cancels out

n = 5

check:
3 times 5 minus 2 equals 13
3   × 5 - 2 = 13
15 - 2 = 13
13 =    13

The number is 5

3 0

3 years ago

Write the equation in standard form for the circle with radius 9 centered at the origin.

amid [387]

The standard form of the equation of a circle of radius r, with (assuming centre h, k) is given as:

(X-h)^2 + (y-k)^2 = r^2

As we are required to write an equation in standard form for the circle with radius 9 centred at the origin.
Centre(h,k)=(0,0), r=9

Substituting these values into the standard form of the equation of a circle given above:

(X-0)^2 + (y-0)^2 = 9^2
X^2 + y^2 =81

The standard form is x^2 + y^2 =81

I’m pretty sure this is right

8 0

2 years ago

If an open box has a square base and a volume of 115 in.3 and is constructed from a tin sheet, find the dimensions of the box, a

Karolina [17]

Let h = height of the box,
x = side length of the base.

Volume of the box is $V=x^{2} h = 115$ .
So $h = \frac{115}{ x^{2} }$

Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
$S = 2( x^{2} + hx + hx) \\ = 2 x^{2} + 4hx \\ = 2 x^{2} + 4( \frac{115}{ x^{2} } )x$
$= 2 x^{2} + \frac{460}{x}$
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.

$S' = 4x- \frac{460}{ x^{2} } = 0$
$4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86$
Then $h= \frac{115}{4.86^{2}} = 4.87$
So the box is 4.86 in. wide and 4.87 in. high.

5 0

3 years ago

Read 2 more answers

The perimeter of the triangular park shown on the right is 10x - 3. What is the missing length?

viva [34]

Answer:

6x - 1

Step-by-step explanation:

Okay this is basically polynomial algebra:

1. Add the value of the lengths already given.

like terms: 3x + x = 4x, -4 + 2 = -2. Now you're left with, 4x - 2.

2. Find the difference between the perimeter and the sum of the lengths.

(10x -3) + (4x + 2) = 6x - 1

4 0

3 years ago