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3 years ago

Solve each equation and check your solution:

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

5 0

Answer:

A. w = 26/2 = -13

B. 3n = 20

n = 20/3

C. -w = 30

w = -30

D. -2c = 32

c = -16

E. 4h = -11

h = -11/4

F. 3.7x = 7.5

x = 75/37

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Given: △DMN, DM=10 3 m∠M=75°, m∠N=45° Find: Perimeter of △DMN

OlgaM077 [116]

Angle D is 180° -75° -45° = 60°. Drawing altitude MX to segment DN divides the triangle into ΔMDX, a 30°-60°-90° triangle, and ΔMNX, a 45°-45°-90° triangle. We know the side ratios of such triangles (shortest-to-longest) are ...

... 30-60-90: 1 : √3 : 2

... 45-45-90: 1 : 1 : √2

The long side of ΔMDX is 10√3, so the other two sides are

... MX = MD(√3/2) = 15

... DX = MD(1/2) = 5√3

The short side of ΔMNX is MX = 15, so the other two sides are

... NX = MX(1) = 15

... MN = MX(√2) = 15√2

Then the perimeter of ΔDMN is ...

... P = DM + MN + NX + XD

... P = 10√3 +15√2 + 15 + 5√3

... P = 15√3 +15√2 +15 . . . . perimeter of ΔDMN

5 0

3 years ago

Are these numbers divisible by 2,3,5,10 444: 90: 45:

anyanavicka [17]

444 ÷ 2 = 222

444 ÷ 3 = 148

90 ÷ 2 = 45

90 ÷ 3 = 30

90 ÷ 5 = 18

90 ÷ 10 = 9

45 ÷ 3 = 15

45 ÷ 5 = 9

So 444 can be divided into 2 and 3

90 can be divided into 2, 3, 5, and 10

45 can be divided into 3 and 5

Hope this helps. :)

6 0

3 years ago

Read 2 more answers

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

What is the sum of the first 11 terms of the arithmetic series in which a1= -12 and d=5?

SVEN [57.7K]

Hello :
the terme general is : an = a1 +(n-1)d
a11 = a1+(11-1)×5
a11 = -12+50
a11 = 38
the sum of the first 11 terms is : S11 = (11/2)(a1+a11)
S11 = (11/2)(-12+38)
S11 = 143

3 0

3 years ago

Insert commas suitably and write the names according to International system of numeration: a) 40050017 b) 251400

kobusy [5.1K]

Answer:

a) 40050017 = 40,050,017

Forty million fifty thousand and seventeen

b) 251400 = 251,400

Two hundred fifty one thousand and four hundred

Step-by-step explanation:

In international system commas are placed after three digits starting from the unit's place.

a) 40050017 = 40,050,017

Forty million fifty thousand and seventeen

b) 251400 = 251,400

Two hundred fifty one thousand and four hundred

It can be placed as shown below.

TM, M, HTh , TTh, Th, H, T, O

4 0 0 5 0 0 1 7

2 5 1 4 0 0

3 0

2 years ago