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Harman [31]
2 years ago
8

Darrin graphs the following inequality: y > -3x -5. Does the side of the line with all solutions (the shaded region) include

the test point (0,0)?
Mathematics
1 answer:
Anarel [89]2 years ago
3 0

Yes it does include the test point (0, 0)

  • equation:  y > -3x -5

<h3>Find y-intercept:</h3>
  • y > -3(0) -5
  • y > -5

<h3>Find x-intercept:</h3>
  • y > -3x -5

<u>switch sides</u>

  • -3x - 5 < 0
  • -3x < 5

<u>Multiply both sides by -1 reverses the inequality</u>

  • x > -5/3

Thus, the following (0, 0) comes under the following x > -5/3 and y > -5

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Solve for x in the equation 3 x squared minus 18 x + 5 = 47.
ryzh [129]

Answer:

x= 3 +\sqrt{23}, 3 -\sqrt{23}

Step-by-step explanation:

You use the quadratic formula to get x= \frac{18+6\sqrt{23} }{6} ,\frac{18-6\sqrt{23} }{6}

Then you simplify and get the answers x= 3 +\sqrt{23}, 3 -\sqrt{23}

​​  

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6 0
3 years ago
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What is the domain of the function y = log(x + 3)?
nata0808 [166]

Find the domain of

y = log(x + 3)

Logarithms can only be taken for positive numbers. So you must have

x + 3 > 0

x > – 3          ✔

So the domain of the function is

D = {x ∈ R:  x > – 3}

or using the interval notation

D = (– 3, +∞)


I hope this helps. =)

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4 years ago
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Find sin 2x, cos 2x, and tan 2x from the given information.<br> cscx=4,tanx&lt;0.
lutik1710 [3]

Answer:

Step-by-step explanation:

Given the expression cosec (x) = 4 and tan(x)< 0

since cosec x = 1/sinx

1/sinx = 4

sinx = 1/4

From SOH, CAH TOA

sinθ = opposite/hypotenuse

from sinx = 1/4

opposite = 1 and hypotenuse = 4

to get the adjacent, we will use the Pythagoras theorem

adj² = 4²-1²

adj² = 16-1

adj ²= 15

adj = √15

cosx = adj/hyp = √15/4

tanx = opposite/adjacent = 1/√15

since tan < 0, then tanx = -1/√15

From double angle formula;

sin2x = 2sinxcosx

sin2x = 2(1/4)(√15/4)

sin2x = 2√15/16

sin2x = √15/8

for cos2x;

cos2x = 1-2sin²x

cos2x = 1-2(1/4)²

cos2x = 1-2(1/16)

cos2x= 1-1/8

cos2x = 7/8

for tan2x;

tan2x  = tanx + tanx/1-tan²x

tan2x = 2tanx/1-tan²x

tan2x = 2(-1/√15)/1-(-1/√15)²

tan2x = (-2/√15)/(1-1/15)

tan2x = (-2/√15)/(14/15)

tan2x = -2/√15 * 15/14

tan2x = -30/14√15

tan2x = -30/7√15

rationalize

tan2x =  -30/7√15 * √15/√15

tan2x =  -30√15/7*15

tan2x =  -2√15/7

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The answer seems to be C

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