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2 years ago

Carrie is seven years older than Dan in four years the sum of their ages will be 63 find the age of each man now

Mathematics

1 answer:

Ksenya-84 [330]2 years ago

4 0

never mind my answers wrong

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A pilot needs to know the maximum height an aircraft can fly at. The cabin has

oee [108]

round it to a whole number of 100 m witch i think would be 15700 m

also i dint realize it said round it to a whole number and to 100 m so it would also be a whole number of 16000 i think if not this than 20000

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3 years ago

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Solve d = rt for r when d = 268mi and t=4hr

vichka [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Find the missing dimension of the prism. Round your answer to the nearest hundredth.

Nitella [24]

Answer:

$x = 6.99in$

Step-by-step explanation:

Given

$l =2in$

$Volume = 72.66in^3$

See attachment for prism

Required

Find x

The prism has a regular hexagon base

The base area is calculated as:

$Base\ Area= \frac{3\sqrt 3}{2}l^2$

This gives:

$Base\ Area= \frac{3\sqrt 3}{2} * 2^2$

$Base\ Area= \frac{3\sqrt 3}{2} * 4$

$Base\ Area= 3\sqrt 3 * 2$

$Base\ Area= 6\sqrt 3$

Side x represents the height.

So, we have:

$Volume = Base\ Area*x$

Make x the subject

$x = \frac{Volume }{ Base\ Area}$

$x = \frac{72.66in^3}{6\sqrt 3 in^2}$

$x = \frac{72.66}{6\sqrt 3}in$

$x = \frac{12.11}{\sqrt 3}in$

$x = 6.99in$ --- approximated

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2 years ago

In which quadrant would (8, -18) be in?

docker41 [41]

It would be in the fourth quadrant because x is positive and y is negative. The quadrants are labeled in counter-clockwise order, starting in the upper-right.

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2 years ago

A line passes through the points 3,10 and 6, 16 write the equation of this line in slope-intercept form

cluponka [151]

$(\stackrel{x_1}{3}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{16}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{16}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{3}}} \implies \cfrac{ 6 }{ 3 } \implies 2$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{ 2}(x-\stackrel{x_1}{3}) \\\\\\ y-10=2x-6\implies {\Large \begin{array}{llll} y=2x+4 \end{array}}$

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11 months ago