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3 years ago

Sides KM and FH in the triangles below will be placed together to form a quadrilateral.

Mathematics

1 answer:

Tanzania [10]3 years ago

8 0

Answer:

Parallelogram (A)

Question:

Sides KM and FH in the triangles below will be placed together to form a quadrilateral.

Triangle M L K. Side M L has a length of 15 and side L K has a length of 35. Angle L is 110 degrees. Triangle F G H. Side F G has a length of 35 and side G H has a length of 15. Angle G is 110 degrees.

Which best describes the quadrilateral that will be formed?

parallelogram

rectangle

rhombus

trapezoid

Step-by-step explanation:

Given:

∆MLK:

Side ML = 15

Side LK = 35

Angle L = 110°

∆ FGH:

Side FG = 35

Side GH = 15

Angle G = 110°

Side MK and FH placed together to form a quadrilateral.

A quadrilateral is a polygon which has 4 sides.

See attachment for diagram

From the diagram and information given:

LK is parallel to FG

ML is parallel to GH

MK = FH

∠L = ∠G (opposite angles are congruent)

Since two pairs of opposite sides are parallel and opposite angles are congruent, it is a paralellogram.

A parallelogram is a quadrilateral which has pairs of opposite sides are parallel and equal.

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A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich

Bess [88]

Answer:

a). k = 2.0794

b). $A_{t}=A_{0}(2^{3t} )$

c). 13107200 cells

d). rate of growth after 6 hours = 27255112

e). 4.76 hours

Step-by-step explanation:

From the formula of bacterial population,

$A_{t}=A_{0}e^{kt}$

Where $A_{t}$ = Bacterial population after time t

$A_{0}$ = Initial population

k = relative growth factor

t = duration or time

a). For $A_{t}=2\times 50=100$ cells

and $A_{0}=50$ cells

Time t = 20 minutes = $\frac{1}{3}$ hours

Now we plug in these values in the formula,

100 = $50(e^{\frac{k}{3}})$

$e^{\frac{k}{3}}=2$

By taking natural log on both the sides of the equation,

$\frac{k}{3}(lne)=ln(2)$

k = 3ln(2) = 2.0794

b). To get the expression we will plug in the value of k in the formula.

$A_{t}=A_{0}e^{kt}$

Since k = 3ln(2)

$A_{t}=A_{0}e^{3t(ln2)}$

Let y = $e^{3t(ln2)}$

By taking natural log on both the sides of the equation,

lny = $ln(e^{3t(ln2)} )$

lny = 3t(ln2)ln(e)

ln(y) = 3t(ln2)

ln(y) = $ln(2)^{3t}$

y = $2^{3t}$

Now our expression will be $A_{t}=A_{0}(2^{3t} )$

c). Number of cells after t = 6 hours

$A_{6}=50(2^{3\times 6} )$

$A_{6}=50(2^{18})$

$A_{6}=50\times (262144)=13107200$

d). We can get the rate of growth by finding derivative of the expression

$\frac{d}{dt}(A_{t})=\frac{d}{dt}[A_{0}(e^{kt}})]$

= $A_{0}[ke^{kt}]$

Now $\frac{d}{dt}(A_{6})=k.A_{6}$

= 2.0794×13107200

= 27255112

e) From the given formula,

$A_{t}=A_{0}(2^{3t} )$

$1000000=50(2^{3t} )$

$2^{3t}=20000$

3t(ln2) = ln(20000)

t = $\frac{ln(20000)}{3(ln2)}$

t = $\frac{9.90349}{2.07944}=4.76$ hours

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Answer:

72 = 30%

Step-by-step explanation:

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