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eimsori [14]
2 years ago
7

What is the domain

Mathematics
2 answers:
xenn [34]2 years ago
8 0
<h3>Answer: Give the domain and the range of each quadratic function whose graph is described. The vertex is (−1,−2)(−1,−2) and the parabola opens up.</h3>

elena-14-01-66 [18.8K]2 years ago
3 0
The range because it’s the smallest and the largest number subtracted
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Determine the domain and the range for this relationship.
Vladimir [108]

Answer:

domain 0 ≤ y ≤ 72 and the range is 0 ≤ y ≤ 6

Step-by-step explanation:

I hope this helps you.

7 0
2 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
Math escape room maze ( I need help) I know it goes A,E,I but I don’t know what goes next)
miskamm [114]

Answer:

9

Step-by-step explanation:

Use angle summation:

∠HGE + ∠EGF = ∠HGF

3x + 11 + 110 = 16x + 4

117 = 13x

x = 9

4 0
3 years ago
HELPPPPP Find the missing parts of the triangle. Round to the nearest tenth when necessary or to the nearest minute as appropria
Romashka-Z-Leto [24]

The missing parts of the triangle ABC are A = 31.4°, B = 57.4°, C = 91.2°

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that a = 8.6 in, b = 13.9 in. and c = 16.5 in. Using cosine rule:

c² = a² + b² - 2abcosC

16.5² = 8.6² + 13.9² - 2(8.6)(13.9) * cosC

C = 91.2°

Also:

a² = b² + c² - 2bccosA

8.6² = 16.5² + 13.9² - 2(16.5)(13.9) * cosA

A = 31.4°

A + B + C = 180° (angles in a triangle)

31.4 + B + 91.2 = 180

B = 57.4°

The missing parts of the triangle ABC are A = 31.4°, B = 57.4°, C = 91.2°

Find out more on equation at: brainly.com/question/2972832

#SPJ1

3 0
2 years ago
Me need help on math
SSSSS [86.1K]
Thank you for the free points!!!
7 0
3 years ago
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