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2 years ago

Can someone help me with this please

Mathematics

1 answer:

zhuklara [117]2 years ago

6 0

The error is when she added 4 to both sides.

Instead, she should divide both sides by -4 to undo the multiplication. Specifically, the multiplication between the -4 term and the (x-2)^2 term.

This is one possible correction of what her steps could look like.

$y = -4(x-2)^2\\\\\\0 = -4(x-2)^2\\\\\\\frac{0}{-4} = \frac{-4(x-2)^2}{-4}\\\\\\0 = (x-2)^2\\\\\\(x-2)^2 = 0\\\\\\\sqrt{(x-2)^2} = \sqrt{0}\\\\\\x-2 = 0\\\\\\x-2+2 = 0+2\\\\\\x = 2\\\\\\$

In the third step, I'm dividing both sides by -4. In the sixth step, I'm applying the square root to both sides to undo the squaring operation. Then two steps later, I'm adding 2 to both sides to undo the "-2", which helps fully isolate x.

The only solution is x = 2, so it's the only x intercept. You can verify this answer by plugging x = 2 back into the original equation at the very top. You should end up with y = 0.

Note: The term "root" is another name for "x intercept".

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Find the slope and reduce of p=(3/4,1 1/4) q=(-1/2,-1)

sergij07 [2.7K]

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
p&({{ \frac{3}{4}}}\quad ,&{{ \frac{1}{4}}})\quad 
% (c,d)
q&({{ -\frac{1}{2}}}\quad ,&{{ -1}})
\end{array}$

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-1-\frac{1}{4}}{-\frac{1}{2}-\frac{3}{4}}\quad \cfrac{\leftarrow LCD=4}{\leftarrow LCD=4}
\\\\\\
\cfrac{\frac{-4-1}{4}}{\frac{-2\cdot 1-3}{4}}\implies \cfrac{\frac{-5}{4}}{\frac{-5}{4}}\implies \cfrac{-5}{4}\cdot \cfrac{4}{-5}\implies 1$

7 0

3 years ago

What is the equation of the line passing through the points (–25, 50) and (25, 50) in slope-intercept form?

allochka39001 [22]

The equation of the line passing through the points (-25, 50) and (25, 50) in slope-intercept form is y = 50. Hence, 4th option is the right choice.

The slope-intercept form of a line is written as y = mx + b, where m is the slope of the line, and b is the y-intercept.

The slope of a line passing through the points (x₁, y₁) and (x₂, y₂) can be calculated using the formula, slope (m) = (y₂ - y₁)/(x₂ - x₁).

Therefore, slope of the line passing through the points (-25, 50) and (25, 50) can be calculated as m = (50 - 50)/(25 - (-25)) = 0/(-50) = 0.

We can find the equation of the line using the point-slope formula, according to which, a line having a slope m and passing through the point (x₁, y₁) can be written as y - y₁ = m(x - x₁).

Therefore, the equation of the given line can be written as:

y - 50 = 0(x - 25)

or, y - 50 = 0,

or, y = 50.

Therefore, the equation of the line passing through the points (-25, 50) and (25, 50) in slope-intercept form is y = 50. Hence, 4th option is the right choice.

Learn more about the equation of a line at

brainly.com/question/18831322

#SPJ10

3 0

2 years ago

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N