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Amanda [17]
3 years ago
11

"Write a set of inequalities and solve to find the range of possibilities for x. You will need to write two separate inequalitie

s instead of one compound inequality."

Mathematics
1 answer:
san4es73 [151]3 years ago
3 0

Answer:

  • x > 5/3
  • x < 8

Step-by-step explanation:

<u>First, the side lengths should be greater than zero:</u>

  • 3x - 5 > 0
  • 2x + 3 > 0

Second, The triangles have a common side and the given sides are opposite of 27° and 37° angles.

<u>This tells us the side opposite of greater angle is greater:</u>

  • 2x + 3 > 3x - 5

<u>Simplify each inequality:</u>

  • 3x - 5 > 0 ⇒ 3x > 5 ⇒ x > 5/3
  • 2x + 3 > 0 ⇒ 2x > - 3 ⇒ x > - 3/2
  • 2x + 3 > 3x - 5 ⇒ 3x - 2x < 3 + 5 ⇒ x < 8

<u>Common solution is:</u>

  • x > 5/3
  • x < 8
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Use the function below to find F(4)<br><br>F(x)=5•(1/3)^x
elena-s [515]

Answer:

\frac{5}{81}

Step-by-step explanation:

To evaluate f(4) substitute x = 4 into f(x)

f(4) = 5 × (\frac{1}{3 )} ^{4} = 5 × \frac{1}{3^{4} } = \frac{5}{81}

4 0
4 years ago
Each spring, Rachel starts sneezing when the pear trees on her street blossom. She reasons that she is allergic to pear trees. F
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Answer:

Seasonal allergies

Step-by-step explanation:

She migtht have seaonal allergies

3 0
3 years ago
Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π
Natasha2012 [34]
Let H denote the helicoid parameterized by

\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k

for 0\le u\le1 and 0\le v\le9\pi. The surface area is given by the surface integral,

\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv

We have

\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j
\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k
\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k
\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}

So the area of H is

\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv
=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2
5 0
3 years ago
Which graph represents the function of f(x) = the quantity of 9 x squared minus 36, all over 3 x plus 6?
alexira [117]
F ( x ) = ( 3 x + 6 ) ( 3 x - 6 ) / ( 3 x + 6 ) = 3 x - 6
and for domain : 3 x + 6  ≠ 0
3 x ≠ - 6
x ≠ - 2
Answer:
C ) graph of 3 x - 6, with discontinuity at - 2.
4 0
3 years ago
Read 2 more answers
What is the approximate area of the circle shown below?
bekas [8.4K]

Answer:

A. 804 cm²

Step-by-step explanation:

Formula :

\boxed{A = \pi r^2}

A = 3.14 × 16²

A = 3.14 × (16 × 16)

A = 3.14 × 256

A = 803.84 cm²

A = <u>8</u><u>0</u><u>4</u><u> </u><u>cm²</u> (rounded)

So, the approximate area of the shown below is 804 cm² (A)

#CMIIW ^^

8 0
3 years ago
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