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3 years ago

For which would you draw a solid boundary line and shade to the right?

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

Answer:

x ≤ -3

Step-by-step explanation:

For solid lines you would do ≤ or ≥, then for shading on the right side you would need < or ≤

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A researcher says to the respondents in a poll, “Eating too many sugary foods leads to cavities. Would you rather have soda or w

photoshop1234 [79]

Answer:

. D. Yes, the more information provided by a researcher the better. Respondents can now give an informed opinion and the results will be more accurate.

Step-by-step explanation:

But again this could be an opinion answer as well

Hope this helps

If this seems incorrect anyway please just comment and I shall change my answer thanks very much :)

6 0

3 years ago

A conservative estimate of the number of stars in the universe is 6 x 10^22. The average human can see about 3,000 stars at nigh

Ilia_Sergeevich [38]

$2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

<h3><u>Solution:</u></h3>

Given that, conservative estimate of the number of stars in the universe is $6 \times 10^{22}$

The average human can see about 3,000 stars at night with only their eyes

To find: Number of times more stars are there in the universe, compared to the stars a human can see

Let "x" be the number of times more stars are there in the universe, compared to the stars a human can see

Then from given statement,

$\text{Stars in universe} = x \times \text{ number of stars human can see}$

<em><u>Substituting given values we get,</u></em>

$6 \times 10^{22} = x \times 3000\\\\x = \frac{6 \times 10^{22}}{3000}\\\\x = \frac{6 \times 10^{22}}{3 \times 10^3}\\\\x = 2 \times 10^{22-3}\\\\x = 2 \times 10^{19}$

Thus $2 \times 10^{19}$ times more stars are there in universe compared to human eye can see

5 0

4 years ago

Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral

MAVERICK [17]

Answer:

a. $\frac {2} {s-1}$ converges to s> 1.

b. $\frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $- \frac {2}{s + 3}$ converges to s> - 3.

d. $\frac {s}{s^2 + 25}$ converges to s> 0.

e. $\frac {10} {s^2 + 1}$ converges even s> 0.

f. $\frac {12}{s^2 + 4}$ converges to s> 0.

g. $-\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $\frac {1} {s ^ 2 + 4}$ converges to s> 0.

Step-by-step explanation:

a. $L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1}$ converges to s> 1.

b. $L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3}$ converges to s> - 3.

d. $L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25}$ converges to s> 0.

e. $L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1}$ converges even s> 0.

f. $L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4}$ converges to s> 0.

g. $L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4}$ converges to s> 0.

7 0

4 years ago

Find the value of 5x - 2y+ 3z when x = 7 , y = 11 and z = -1

Oksanka [162]

By replacing you get:

5*7-2*11+3*(-1) = 35-22-3= 10

so the answer is 10.

If you like my answers, feel free to follow me and ask me more maths problems. :)

7 0

3 years ago

Read 2 more answers

Express 5.1146• 10^3 in standard notation

charle [14.2K]

Answer:

5,114.6

Step-by-step explanation:

5.1146 x 10^3 means move decimal point 3 to the right to get the answer

4 0

3 years ago