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3 years ago

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What is an equation in point-slope form for the line that passes through the points (−3,5) and (2,−3)?

1 answer:

fredd [130]3 years ago

4 0

(-3,5). (2,-3)
x1,y1. x2,y2

y2-y1. -3-5. -8
--------- = ------- = ----- =8 =m
x2-x1. 2-3. -1

plug the rest in

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2logx=3-2log(x+3) solve for x

AVprozaik [17]

Answer:

$\large\boxed{x=\dfrac{-3+\sqrt{40+10\sqrt{10}}}{2}}$

Step-by-step explanation:

$2\log x=3-2\log(x+3)\\\\Domain:\ x>0\ \wedge\ x+3>0\to x>-3\\\\D:x>0\\============================\\2\log x=3-2\log(x+3)\qquad\text{add}\ 2\log(x+3)\ \text{to both sides}\\\\2\log x+2\log(x+3)=3\qquad\text{divide both sides by 2}\\\\\log x+\log(x+3)=\dfrac{3}{2}\qquad\text{use}\ \log_ab+\log_ac=\log_a(bc)\\\\\log\bigg(x(x+3)\bigg)=\dfrac{3}{2}\qquad\text{use the de}\text{finition of a logarithm}\\\\x(x+3)=10^\frac{3}{2}\qquad\text{use the distributive property}$

$x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D$

6 0

3 years ago

Shane and Abha earned a team badge that required their team to collect no less than 2000 cans for recycling. Abha collected 178

Katarina [22]

Answer:

$2x+178\geq 2000$

Step-by-step explanation:

Let, the number of cans collected by Shane = x.

So, the number of cans collected by Abha = x + 178.

Since, at least 2000 cans are required to be collected.

Thus, we have the inequality,

Number of cans by Shane + Number of cans by Abha ≥ 2000.

i.e. $x+(x+178)\geq 2000$

i.e. $2x+178\geq 2000$

Thus, the required inequality is $2x+178\geq 2000$ .

4 0

3 years ago

What expression represents six times a number plus 8 times a number

zheka24 [161]

6n + 8n

Which is 6 * n + 6 * 8, n being a number.

4 0

3 years ago

What is the variable y in w= x y/z?

AleksAgata [21]

Multiply by z to get
wz = xy
divide by x
y = wz/x

8 0

3 years ago

A 748-N man stands in the middle of a frozen pond of radius 4.0 m. He is unable to get to the other side because of a lack of fr

Ber [7]

Answer:

The man will take 64 seconds to reach to the south shore of the frozen pond.

Step-by-step explanation:

Given:

Weight of the man = 748 N Mass of the man, $(m)= \frac{W}{g}$ = $\frac{748}{9.8}$ = $76.32$ kg

Radius of the pond $(r)$ = 4 m

Mass of the textbook = 1.2 kg

Velocity at which the textbook is thrown = 4 ms^1

We have to find the velocity of the man after the throw.

Let the velocity is $V_m$ .

Now using law of conservation of momentum we can find the $V_m$ value.

$m_(_b_)V_b_(_i_) +m_(_m_)V_m_(_i_) =m_(_b_)V_b_(_f_)+m_(_m_) V_m_(_f_)$

Considering $V_m_(_f_)=V_m$

And initial velocity of both the man and book i.e $V_b_(_i_)=0,\ V_m_(i_)=0$

So,

⇒ $0 =m_(_b_)V_b_(_f_)+m_(_m_) V_m$

⇒ Plugging the values.

⇒ $V_m=-\frac{m_(_b_)V_b_(_f_)}{m_(_m_)}$

⇒ $V_m=-\frac{1.2\times 4}{76.32}$

⇒ $V_m=-0.062$ ms^-1

Here the negative velocity is meant for opposite direction of the throw.

Numerically we will write, $V_m = 0.062$

With this velocity the man will move towards south.

We have to calculate the time taken by the man to move to its south shore.

And we know $velocity(v)\times time(t) = distance(d)$

Let the time taken be $t$ and $v\times t = d$ and $d=r$ then, $V_m\times t=r$

Then

⇒ $t=\frac{radius\ (r)}{V_m}$

⇒ Plugging the values.

⇒ $t=\frac{4}{0.062}$

⇒ $t =64$ sec

The man will take 64 seconds to reach to the south shore of the frozen pond (circular).

5 0

3 years ago