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3 years ago

Solve for x 5^x= 3x^x+1

Mathematics

1 answer:

Stella [2.4K]3 years ago

3 0

Answer:

x=1

Step-by-step explanation:

5 − 3 x = x + 1

Move all terms containing x to the left side of the equation.

5 − 4 x = 1

Move all terms not containing x to the right side of the equation. − 4 x = − 4

Divide each term by − 4 and simplify.

x = 1

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No work needed!! Just answers, thank you! 25 points!!

Maurinko [17]

Answer:

Step-by-step explanation:

$S_{31} = 9 \cdot 31 + [-0.5 \cdot \frac{31(31-1)}{2}] = 46.5$

$S_{40} = 40 + [-3 \cdot \frac{40(40-1)}{2}] = -2300$

3 0

3 years ago

Find the equation of the tangent line to the curve (a lemniscate)

olya-2409 [2.1K]

Answer:

$m=\frac{9}{13}$ and $b=\frac{40}{13}$

Step-by-step explanation:

The equation of curve is

$2(x^2+y^2)^2=25(x^2-y^2)$

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

$2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})$

$4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})$

The point of tangency is (-3,1). It means the slope of tangent is $\frac{dy}{dx}_{(-3,1)}$ .

Substitute x=-3 and y=1 in the above equation.

$4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})$

$40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})$

$-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}$

$80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240$

$130\frac{dy}{dx}=90$

Divide both sides by 130.

$\frac{dy}{dx}=\frac{9}{13}$

If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

$y-y_1=m(x-x_1)$

The slope of tangent line is $\frac{9}{13}$ and it passes through the point (-3,1). So, the equation of tangent is

$y-1=\frac{9}{13}(x-(-3))$

$y-1=\frac{9}{13}(x)+\frac{27}{13}$

Add 1 on both sides.

$y=\frac{9}{13}(x)+\frac{27}{13}+1$

$y=\frac{9}{13}(x)+\frac{40}{13}$

Therefore, $m=\frac{9}{13}$ and $b=\frac{40}{13}$ .

5 0

2 years ago

Write the slope-intercept form of the equation that passes through the point (-3, 5) and is perpendicular to the line y = 1/5x +

aleksley [76]

Answer:

Step-by-step explanation:

$\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\=========================$

$\text{We have}\ y=\dfrac{1}{5}x+10\to m_1=\dfrac{1}{5}\\\\\text{Therefore}\ m_2=-\dfrac{1}{\frac{1}{5}}=-5.\\\\\text{Put the value of a slope and the coordinates of the point (-3, 5)}\\\text{to the equation}\ y=mx+b:\\\\5=-5(-3)+b\\5=15+b\qquad\text{subtract 15 from both sides}\\-10=b\to b=-10\\\\\text{Finally:}\\\\y=-5x-10$

7 0

3 years ago

A selective university advertises that 96% of its bachelor’s degree graduates have, on graduation day, a professional job offer

OLEGan [10]

Answer:

The probability is $P( p < 0.9207) = 0.0012556$

Step-by-step explanation:

From the question we are told

The population proportion is $p = 0.96$

The sample size is $n = 227$

The number of graduate who had job is k = 209

Generally given that the sample size is large enough (i.e n > 30) then the mean of this sampling distribution is

$\mu_x = p = 0.96$

Generally the standard deviation of this sampling distribution is

$\sigma = \sqrt{\frac{p (1 - p )}{n} }$

=> $\sigma = \sqrt{\frac{0.96 (1 - 0.96 )}{227} }$

=> $\sigma = 0.0130$

Generally the sample proportion is mathematically represented as

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{209}{227}$

=> $\^ p = 0.9207$

Generally probability of obtaining a sample proportion as low as or lower than this, if the university’s claim is true, is mathematically represented as

$P( p < 0.9207) = P( \frac{\^ p - p }{\sigma } < \frac{0.9207 - 0.96}{0.0130 } )$