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sertanlavr [38]
3 years ago
13

Which inequality represents the expression 9 is less than or equal to b

Mathematics
2 answers:
Mila [183]3 years ago
6 0

Answer:

9< with the line under the symbol then you put 'b'

Step-by-step explanation:

brilliants [131]3 years ago
5 0

Answer: The answer is 9 ≤ b

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Can someone help with this question?
elixir [45]
When you subtract flip the signs of the second set.
2x^2+5-4x+3

Now combine like terms
2x^2-4x+8

Final answer: B
5 0
3 years ago
Read 2 more answers
Find tan(30) (its a right triangle)
Travka [436]
Use the 30-60-90 triangle.

Tan is opposite over adjacent.

\frac{1}{ \sqrt{3} }
Square roots cannot be on the bottom.

\frac{1}{ \sqrt{3} }  \times  \frac{ \sqrt{3} }{ \sqrt{3} }
The bottom cancel out the square root bar.

Your answer is:

\frac{ \sqrt{3} }{3}
(It's best to memorize the 30-60-90 and the 45-45-90 triangles during trigonometry. It helps a lot.)

Hope this helps!

:)

8 0
3 years ago
kelly has 420ml of lime juice and an unlimited supply of lemon and orange juice what is the maximum amount of fruit drink kelly
Sliva [168]
  1. Kelly can make an unlimited amount of fruit drink
3 0
3 years ago
Ms. Albero decided to make juice to serve along with the pizza at the Student Government party. The directions said to mix 3 sco
mixas84 [53]

Answer: ok

Step-by-step explanation:

3 0
3 years ago
Solve the initial value problems:<br> 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1
ladessa [460]

Answer:

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi  - \frac{1}{2}

Step-by-step explanation:

Given the initial value problem \frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits  \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi  = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi  - \frac{1}{2}

The solution to the initial value problem will be;

-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi  - \frac{1}{2}

5 0
3 years ago
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