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2 years ago

16. What are the values of a and b?

Mathematics

1 answer:

MrMuchimi2 years ago

4 0

Answer:

a = 20; b = 2√101

Step-by-step explanation:

The triangle at upper left and the one at lower right are similar. Ratios of corresponding sides are equal, so ...

200/a = a/2

400 = a^2

√400 = 20 = a

Then, by the Pythagorean theorem, the value of b can be found:

b = √(a^2 + 2^2) = √(400 +4) = √4·√101

b = 2√101

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Find the missing measures. Write your numerical answers only with a comma in between.

Katen [24]

Answer:

m∠T = 72°

m∠U = 54°

Step-by-step explanation:

Property of an isosceles triangle,

Opposite angles of two equal sides of an isosceles triangle are equal in measure.

By this property, in the given triangle STU,

Since, ST ≅ TU,

m∠S = m∠U = 54°

Since, sum of interior angles of a triangle is 180°

m∠T + m∠S + m∠U = 180°

m∠T + 54° + 54° = 180°

m∠T = 180° - 108°

m∠T = 72°

7 0

2 years ago

A train covers a distance of 80 miles in 2 hours. At this rate, about how many hours will it take to cover a distance of 200 mil

Vadim26 [7]

Let's write miles over hours
80 / 2 = 200/ x
cross multiply
80x = 400
divide both sides by 80
x = 5

It will take 5 hours to cover 200 miles

Hope this helps :)

6 0

3 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

2 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago

Y varies directly as x and y =18 when x=1/2 write an equation that relates y and x

marusya05 [52]

Answer:

y = 36x

Step-by-step explanation:

Given that y varies directly as x, then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 18 when x = $\frac{1}{2}$ , thus

k = $\frac{y}{x}$ = $\frac{18}{\frac{1}{2} }$ = 36

y = 36x ← equation of variation

6 0

2 years ago