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2 years ago

A plane intersects a double-napped cone such that the plane contains the generating line.

Mathematics

1 answer:

Mice21 [21]2 years ago

7 0

Answer:

line and degenerate parbola

Step-by-step explanation:

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In a 4 day period, a delivery person delivered the following number of

Westkost [7]

Answer: 1400 packages

Step-by-step explanation:

1150+1200+1900+1350 divided by the 4 day interval means and average of:

1400 packages per day

6 0

2 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

The heights of young women aged 20 to 29 follow approximately the N(64, 2.7) distribution. Young men the same age have heights d

murzikaleks [220]

Answer: 10.703%

Step-by-step explanation:

Let minimum height of the tallest 25% of young women be M.

Let Q be the random variable which denotes the height of young women.

Therefore, Q – N(64,2.70)

Now, P(Q˂M) = 1-0.25

i.e. P[(Q-64)/2.7 ˂ (M-64)/2.7] = 0.75

I.e. ф-1 [(M-64)/2.7] = 0.75 i.e. (M-64)/2.7 = ф-1 (0.75) = 0.675 i.e. M = 65.8198 inches

Let R be the random variable denoting the height of young men

Therefore, R – N (69.3, 2.8)

i.e. (R-69.3)/2.8 – N(0,1)

therefore the probability required = P(R ˂65.8198) = P[(R-69.3)/2.8 ˂ (65.8198 – 69.3)/2.8]

this gives P[(R-69.3)/2.8 ˂] = ф (-1.2429) = 0.107033

From this, the percentage of young men shorter than the shortest amongst the tallest 25% of young women is 10.703%

4 0

3 years ago

Please explain how you got the answer.

photoshop1234 [79]

Answer:

Step-by-step explanation:

First, simplify the equation they give you

125x+200≥1200 ---------> 125x≥1000---------->x≥8

So, x has to be greater than or equal to 8 which rules out b or d.

So the answers are narrowed down to a and c, the only difference is that the dot in A is filled in whereas the dot in c is not. In math, having a filled in circle means that value is included (≥ sign), so we will choose the answer a.

8 0

2 years ago

Read 2 more answers

This graph shows the solution to which inequality? A.y>=1/3x-2 B.y<1/3x-2 C.y<=1/3x-2 D.y>1/3x-2

Alina [70]

Answer:

Step-by-step explanation:

3 0

2 years ago