FIND THE VOLUME OF EACH SNOWBALL)

Find an equation of a line that comes close to the points listed in the table.

sergey [27]

$y=x+2\\x=6\to y=6+2=8\\x=10\to y=10+2=12\\x=12\to y=12+2=14\\x=16\to y=16+2=18\\x=18\to y=18+2=20\\-----------------------\\y=x+3\\x=6\to y=6+3=9\\x=10\to y=10+3=13\\x=12\to y=12+3=15\\x=16\to y=16+3=19\\x=18\to y=18+3=21\\-----------------------\\Answer:y=x+1.5$

3 0

2 years ago

HELPPPPPP MEEEE THANKSS

stepan [7]

3,500 it is that your welcome I have the same problem right now

6 0

2 years ago

Read 2 more answers

Find the obtuse angle the line y-2x=7 makes with the x- axis

Kazeer [188]

In analytical geometry, there are already derived equations to find the distance of lines and points as well as the angle made between two lines. As special case is when the other line is one of the coordinate axis. Then, the formula can be simplified to

tan θ =m, where m is the slope of the equation

In the next step, we also incorporate operations of calculus. Since the slope is equal to Δy/Δx, this is equivalent to dy/dx in calculus. Therefore, you can find the slope by differentiating the equation in terms of x.

y-2x=7
y = 2x+7
dy/dx = 2 =m

So,

tan θ = 2
θ = tan⁻¹(2)
θ = 63.43°

5 0

3 years ago

I need help with this please

Doss [256]

Answer:I feel bad

Step-by-step explanation:

3 0

2 years ago

Anyone can help me solve this equation using cross multiplying

Natali5045456 [20]

9514 1404 393

Answer:

x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)$

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

$3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}$

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Comment on "cross multiply"

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes what the result looks like. It doesn't adequately describe how you get there. The one and only rule in solving Algebra problems is "whatever is done to one side of the equation must also be done to the other side of the equation." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "multiply by the product of the denominators and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

$\displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}$

8 0

2 years ago