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3 years ago

There are 40 students in a class.

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

5 0

Step-by-step explanation:

Total no. of students = 40

Total no. of girls = 22

Total no. of boys = 40 - 22 = 18

No. of girls walking to school = 9

No. of girls who cycle to school = 22 - 9

No. of boys who cycle to school = 7

No. of students who take the bus = 10

(6 boys and 4 girls)

No. of boys who walk to school = 18 - (7 + 6) = 5

Total no. of students who walk to school = 9 + 5 = 14

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On circle O below, the measure of FJ is 84º . The measure of GH is 76°.<br> What is the measure of

Vsevolod [243]

Given:

Measure of arc FJ = 84°

Measure of arc GH = 76°

To find:

The measure of angle HKJ.

Solution:

The image of the question is attached below.

Angles inside the circle theorem:

If two chords intersect inside a circle, then the measure of each angle is one half the sum of the measures of the intercepted arc and its vertical arc.

$$\Rightarrow m \angle G K H=\frac{1}{2}(m(a r \ G H)+m(a r \ F J))$

$$\Rightarrow m \angle G K H=\frac{1}{2}\left(76^{\circ}+84^{\circ}\right)$

$$\Rightarrow m \angle G K H=\frac{1}{2}\left(160^{\circ}\right)$

$\Rightarrow m \angle G K H=80^{\circ}$

Sum of the adjacent angles in a straight line is 180°.

⇒ m∠GKH + m∠HKJ = 180°

⇒ 80° + m∠HKJ = 180°

Subtract 80° from both sides.

⇒ 80° + m∠HKJ - 80° = 180° - 80°

⇒ m∠HKJ = 100°

The measure of ∠HKJ is 100°.

8 0

3 years ago

The cost of printing books C varies partly as the amount A charged by the printer

lawyer [7]

The equation connecting C,A and n is C = 25A - $\frac{49990000}{n}$

<h3>What is variation?</h3>

Variation is a mathematical term used to show how a quantity behaves when one quantity increases or decreases.

Analysis:

C = AK + M/n

where K and M are constants

when n = 1000, A = 5000, C = 75010

75010 = 5000K + M/1000-----------1

50004 = 10000k + M/200-------------2

solving simultaneously, K = 25, M = -49990000

C = 25A - 49990000/n

In conclusion, the equation connecting C,A and n is C = 25A - 49990000/n

Learn more about variation: brainly.com/question/6499629

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2 years ago

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3 years ago

The logistic equation for the population (in thousands) of a certain species is given by:

Eva8 [605]

Answer:

b. 1.5

c. 1.5

d. No

Step-by-step explanation:

a. First, let's solve the differential equation:

$\frac{dp}{dt} =3p-2p^2$

Divide both sides by $3p-2p^2$ and multiply both sides by dt:

$\frac{dp}{3p-2p^2}=dt$

Integrate both sides:

$\int\ \frac{1}{3p-2p^2} dp =\int\ dt$

Evaluate the integrals and simplify:

$p(t)=\frac{3e^{3t} }{C_1+2e^{3t}}$

Where C1 is an arbitrary constant

I sketched the direction field using a computer software. You can see it in the picture that I attached you.

b. First let's find the constant C1 for the initial condition given:

$p(0)=3=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-1$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-1 } =\frac{3}{2} =1.5$

c. As we did before, let's find the constant C1 for the initial condition given:

$p(0)=0.8=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=1.75$

Now, let's evaluate the limit:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2+1.75e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}+1.75 } =\frac{3}{2} =1.5$

d. To figure out that, we need to do the same procedure as we did before. So, let's find the constant C1 for the initial condition given:

$p(0)=2=\frac{3e^{0} }{C_1+2e^{0} } =\frac{3}{C_1+2}$

Solving for C1:

$C_1=-\frac{1}{2} =-0.5$

Can a population of 2000 ever decline to 800? well, let's find the limit of the function when it approaches to ∞:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } \\\\Divide\hspace{3}the\hspace{3}numerator\hspace{3}and\hspace{3}denominator\hspace{3}by\hspace{3}e^{3t} \\\\ \lim_{t \to \infty} \frac{3 }{2-0.5e^{-3x} }$

The expression $-e^{-3x}$ tends to zero as x approaches ∞ . Hence:

$\lim_{t \to \infty} \frac{3e^{3t} }{2e^{3t}-0.5 } =\frac{3}{2} =1.5$

Therefore, a population of 2000 never will decline to 800.

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3 years ago

A car travels about 46 feet in one second. At this speed,

spin [16.1K]

I think it is D, 92 seconds. I doubled 46 so I hope I helped :))

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3 years ago