All categories

3 years ago

Please Help!!!

1 answer:

6 0

Let's use the quadratic formula
$\frac{ - b( + - ) \sqrt{ {b}^{2} - 4ac} }{4a}$
a = 1, b = 0, c = -18
$\frac{ - (0)( + - ) \sqrt{ {(0)}^{2} - 4(1)( - 18)} } {4(1)} \\ \frac{( + - ) \sqrt{72} }{4}$
This is now split into two equations:
$\frac{ + \sqrt{72} }{4}$
and
$\frac{ - \sqrt{72} }{4}$
Since $\frac{ \sqrt{72} }{4} \\ = \frac{6 \sqrt{2} }{4} \\ = \frac{3 \sqrt{2} }{2}$
The answer is
$\frac{3 \sqrt{2} }{2} \: and \: \frac{ - 3 \sqrt{2} }{2}$
which means there are 2 real number solutions to the problem: answer B.

You might be interested in

What's halfway between 3100and4000

Anni [7]

3950 is in between
is the answer

4 0

2 years ago

Read 2 more answers

Which graph corresponds with the equation <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B1%7D%7B2%20%7D%20%28x%2B1%29%5E

JulijaS [17]

$\qquad\qquad\huge\underline{{\sf Answer}}$

The given equation is a Quadratic equation, so it's graph must be a parabola, and the Coefficient of x² is positive that means the parabola must have an opening upward.

And 2 is added to the ideal equation, so the parabola must have shift 2 units up from x - axis.

From the above information, we can easily conclude that the Correct representation is done in graph 4

5 0

2 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

2 years ago

Read 2 more answers

Look at this set of 10 numbers:

swat32

In the original set of numbers, the median is six
But, if six was added to the set, then the median would be six
So the median wouldn't decrease at all

8 0

3 years ago

Read 2 more answers

Can I PLEASE get some help?

V125BC [204]

The 1st and 3rd choices are irrational. The other 2 are rational. Irrational numbers are numbers that you can't make a simple fraction out of.

7 0

3 years ago