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3 years ago

ZABD and ZDBC are supplementary angles.

Mathematics

2 answers:

Bess [88]3 years ago

5 0

Answer:

Solution given:

x+110°=180[supplementary]

x=180-110=70°

x=70°

MariettaO [177]3 years ago

3 0

Answer:

x=70°

Step-by-step explanation:

x+110=180 linear pair

x=180-110=70

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Solve the following system of equations using the elimination method. 4x + 2y = 12 4x + 8y = –24

Hunter-Best [27]

Answer:

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Step-by-step explanation:

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2 years ago

If you toss 9 fair coins, in how many ways can you obtain at least one tail?

maks197457 [2]

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3 years ago

Someone help, I suck at math

attashe74 [19]

option A.55°

Answer:

Solution given:

<x+<y=<z

exterior angle of a triangle is equal to the sum of two opposite interior angle

4n-18+n+8=133-6n

5n+6n=133+10

11n=143

n=143/11

n=13

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option A.55°

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3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

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3 years ago

Evaluate the expression when b=2 b·5

Drupady [299]

Answer:

$Solution,\\When,b=2\\Now, b*5=2*5=10$

Step-by-step explanation:

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2 years ago