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4 years ago

I need help with 5 and 6

Mathematics

1 answer:

8_murik_8 [283]4 years ago

7 0

Answer:

w = 6

RS = 13

Step-by-step explanation:

RS + ST = RT

(2w + 1) + (w - 1) = 18

3w = 18

; w = 6

Hence the value of RS = 2(6) + 1

; RS = 13

AC = AB + BC

56 = (x + 16) + (5x + 10)

56 = 6x + 26....then rearrange the equation to form...;

56 - 26 = 6x

30 = 6x...

where...; x = 5

AB = 5 + 16 and BC = 5(5) + 10

;AB = 21 ;BC = 35

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3 years ago

A civil engineer is designing a public parking lot for the new town hall. The number of cars in each row should be six less than

Alexandra [31]

The new parking lot must hold twice as many cars as the previous parking lot. The previous parking lot could hold 56 cars. So this means the new parking lot must hold 2 x 56 = 112 cars

Let y represent the number of cars in each row, and x be the number of total rows in the parking lot. Since the number of cars in each row must be 6 less than the number of rows, we can write the equation as:

y = x - 6 (1)

The product of cars in each row and the number of rows will give the total number of cars. So we can write the equation as:

xy = 112 (2)

Using the above two equations, the civil engineer can find the number of rows he should include in the new parking lot.

Using the value of y from equation 1 to 2, we get:

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6 0

3 years ago

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

3 0

3 years ago

What is the gcf for 84

kati45 [8]

84

84 = 42 x 2

The greatest common factor is 42, because 84/2 = 42, & you want the biggest number possible

Usually i need two or more numbers so.. this is the best i can get out of one number

hope this helps tho :D

7 0

3 years ago