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3 years ago

13

How do You supposed to do this i need many method to do it.

1 answer:

defon3 years ago

4 0

I like the substitution method. Which is when you make one equation equal only x or y and plug it into the other equation)

There is also the graphing method. If you graphed it, it might not be quite as accurate (at least on hand, on computer you would be pretty exact)

Then there is the elimination method. You multiply one of the equations by a coefficient so that you can eliminate x or y from the equation.

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HELP PLSSSS<br> Use the diagram below where ABC = DEF to find the solutions for x and y.

Vadim26 [7]

I don’t think you posted a picture along with this so I can’t answer

8 0

3 years ago

The length of the red line segment is 6

Alexandra [31]

Answer:

What is the question or image?

Step-by-step explanation:

Need it to solve

8 0

3 years ago

What is a trianglular number

grandymaker [24]

Answer:

any of the series of numbers (1, 3, 6, 10, 15, etc.) obtained by continued summation of the natural numbers 1, 2, 3, 4, 5, etc.

6 0

3 years ago

In right trapezoid MNPQ

Gala2k [10]

Answer:

about 252.78 ft

Step-by-step explanation:

Define angle QMP as α. Then ...

MN = 60·sin(α)

NP = 60·cos(α)

area MPN = (1/2)(MN)(NP) = 1800sin(α)cos(α)

__

PQ = 60tan(α)

area MPQ = (1/2)(MP)(PQ) = 1800tan(α)

__

The ratio of areas is 2.5, so we have ...

1800tan(α) = 2.5·1800sin(α)cos(α)

1 = 2.5cos(α)² . . . . . . divide by 1800tan(α)

cos(α) = √0.4 . . . . . . solve for cos(α)

__

Then the perimeter is ...

Perimeter = MN +NP +PQ +QM = 60sin(α) +60cos(α) +60tan(α) +60/cos(α)

= 60(sin(α) +cos(α) +tan(α) +sec(α))

= 60(0.774597 +0.632456 +1.224745 +1.581139)

= 60(4.212936) = 252.776

The perimeter of the trapezoid is about 252.776 feet.

_____

With perhaps a little more trouble, you can find the exact value to be ...

perimeter = (6√10)(7+√6+√15)

3 0

3 years ago

The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experim

mezya [45]

Answer:

(a) $N(t)=10000e^{(\frac{ln5}{10})t }$

(b) 25,000

(c) 4.3068 min.

Step-by-step explanation:

Rate of change in the number of bacteria is proportional to the number present.

Let N is the population of bacteria.

$\frac{dN}{dt}$ ∝ N ⇒ $\frac{dN}{dt}=kt$ { k = proportionality constant}

initial population No. = 10,000

$N(t_{1} )$ = 20,000

and $N(t_{1}+10 )=100,000$

(a) For population growth

$N(t)=N_{0}e^{kt}=10000e^{kt}$

$N(t_1)=10,000e^{kt_1}=20,000$

$e^{kt}=2$

$ln(e^{kt_1})=ln(2)$

$kt_1=ln(2)$

$t_{1}=\frac{ln2}{k}$ ----------(1)

$N(t_1+t_{10})=100,000$

$100,000=10,000e^{k(t_1+10)}$

$10=e^{k(t_1+10)}$

$ln10=ln[e^{k(t_1+10)}]$

$k(t_1+10)=ln10$

$k(t_1)=ln10-10k$

$t_1=\frac{ln10-10k}{k}$ ----------(2)

from equation (1) and (2)

$\frac{ln_2}{k}=\frac{ln10-10k}{k}$

$ln10-ln2=10k$

$k=\frac{ln5}{10}$

so expression will be

$N(t)=10000e^{(\frac{ln5}{10})t }$

(b) for t = 20

$N_{(20)}=10,000e\frac{ln5}{10}\times 20$

= $10,000\times e^{2ln5}$

= 10,000 × 25

= 25,000

(c) Since $t_1=\frac{ln2}{k}$ [from equation (1)]

$=\frac{ln2}{\frac{ln5}{10} }$

$=\frac{ln2}{ln5}\times 10$

= 4.3068

= 4.3068 min.

6 0

3 years ago