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kramer
3 years ago
14

Jocelyn said that the triangle below is not a right triangle. Her work is shown below. Explain what she did wrong and show Jocel

yn the correct solution.
Mathematics
1 answer:
jekas [21]3 years ago
7 0

Jocelyn said that the triangle below is not a right triangle. Her work is shown below. Explain what she did wrong, and show Jocelyn the correct solution.

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F(x) = -4x2 – x - 3<br> Find f(1)
nadezda [96]

Answer:

-8

Step-by-step explanation:

-4 (1)^2 -(1) -3

-4-1-3=-8

5 0
2 years ago
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<h2>Answer:</h2><h2>-4</h2><h2 /><h2>Hope this helps!!</h2>
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3 years ago
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The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

7 0
2 years ago
Solve the initial value problem. 7y′−4y=e−πt2, y(0)=a Let a0 be the value of a for which the transition from one type of long-ru
densk [106]

Answer:

as shown in the attached file

Step-by-step explanation:

The detailed steps and application of differential equation, the use of integrating factor to generate the solution and to solve for the initial value problem is as shown in the attached file.

8 0
2 years ago
Hello need help please with Divide rational and simplify
aliina [53]
Dividing fractions is the same as multiplying by its reciprocal so you can change the equation to this
\frac{ - 14 + 7x}{4}  \times  \frac{4}{4 - 2x}
And you can simplify by eliminating the fours and multiplying straight across
\frac{ - 14 + 7x}{4 - 2x}
You can pull out - 7 from the top and 2 from the bottom so you have this
\frac{ - 7(2  -  x)}{2(2 - x)}
And then you can cancel out to (2-x) to get an answer of
\frac{ - 7}{2}
5 0
2 years ago
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