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3 years ago

9

Compare and order the numbers below.

1 answer:

Leokris [45]3 years ago

6 0

Order
This is the explanation and the answer for order

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3+-√(-3)^2 - 4(5)(-1)

Alona [7]

Answer:

Step-by-step explanation:

Easy way to do this is step by step. Your quadratic, from your entry, must be

$5x^2-3x-1$ .

Step by step looks like this, one thing at a time:

$x=\frac{3+\sqrt{(-3)^2-4(5)(-1)} }{2(5)}$ becomes

$x=\frac{3+\sqrt{9-(-20)} }{10}$ becomes

$x=\frac{3+\sqrt{9+20} }{10}$

and this of course is

$x=\frac{3+\sqrt{29} }{10}$

Do the same with the subtraction sign to get the other solution.

If you're unsure of how to enter it into your calculator, do it step by step so you don't mess up the sign. If you enter it incorrectly, you could end up with an imaginary number when it should be real, or a real one that should be imaginary.

Just my advice as a high school math teacher.

3 0

3 years ago

What is the equation of the horizontal line that passes through the point (-2, -1)

diamong [38]

Answer:

y= -1

Step-by-step explanation:

A horizontal line would mean that it has no slope, since the line is not decreasing or increasing. The horizontal line drawn through (0, -1) would pass through (-2, -1), since they share the same y coordinate.

5 0

3 years ago

Someone please helppp

KiRa [710]

Answer:

2,16 16,2

Step-by-step explanation:

probally wrong

7 0

3 years ago

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Ms. De Leon wants to produce different sets of test questions for her essay test. If she plans to do this by putting together 3

gladu [14]

The correct answer is b

5 0

3 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

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