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Vesna [10]
3 years ago
6

Helpp

Mathematics
1 answer:
timama [110]3 years ago
7 0

Answer:

X² + 1 (B)

Step-by-step explanation:

Y = X² + 1

x =2

then X² + 1 = y

(2)² + 1 = 5 (that is your Y)

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Please help fast ^^
allsm [11]
The discriminante :
b^2-4ac

1^2 - 4 * -2 * -28 = 1 - 224 = -223

When the discriminant (b^2-4ac) is less than 0, the equation had no real solutions.

-223<0, so, 2x^2+x-28 = 0 has no real solutions.


Hope that helps :)
7 0
2 years ago
Which choice is equivalent to the expression below?
Phantasy [73]
Hello! The answer should be B: 6i\sqrt{6 . Since the square root is above a negative, an imaginary number would come out. Because of this, you can easily recognize that it would be B. 
However, if you can't, you could use the calculator to check (ignoring the negatives), 6\sqrt{6 is equal to \sqrt{216}. 
The final answer should be B. 
7 0
3 years ago
Read 2 more answers
Find the sum of the geometric series: <br>32 - 8 + 2 - 1/2 +...
ZanzabumX [31]
Time to work it out 
32 - 8 + 2 - 1/2
32 - 8 + 2 = 26
now you have
-1/2 + 26 
so we convert 26 to fraction form to subtract 
26/1 - 1/2
we must find the common denominator
52/2 - 1/2
51/2
fifty-one over 2 (fraction) is your answer! :)
6 0
3 years ago
How many 100 dollar bills are in 1 thousand
bonufazy [111]
100 goes into 1,000 ten times. The answer is 10. 
5 0
3 years ago
Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin&lt;0
Gre4nikov [31]

Answer:

Using trigonometric ratio:

\sec \theta = \frac{1}{\cos \theta}

\tan \theta = \frac{\sin \theta}{\cos \theta}

From the given statement:

\cos \theta = -\frac{4}{5} and sin < 0

⇒\theta lies in the 3rd quadrant.

then;

\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}

Using trigonometry identities:

\sin \theta = \pm \sqrt{1-\cos^2 \theta}

Substitute the given values we have;

\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}

Since, sin < 0

⇒\sin \theta = -\frac{3}{5}

now, find \tan \theta:

\tan \theta = \frac{\sin \theta}{\cos \theta}

Substitute the given values we have;

\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}

Therefore, the exact value of:

(a)

\sec \theta =-\frac{5}{4}

(b)

\tan \theta= \frac{3}{4}

7 0
3 years ago
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