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asambeis [7]
3 years ago
10

What does the fundamental theorem of algebra state about the equation 2x^2-x+2=0

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

See below

Step-by-step explanation:

The Fundamental Theorem of Algebra states that the degree of the polynomial is the number of roots, where some may be complex depending on the polynomial or degree.

In this case, 2x^2-x+2=0 has its highest degree of 2, so there are 2 roots.

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A rectangle is a _____.<br><br> a. quadrilateral<br> b. square<br> c. parallelogram<br> d. rhombus
Pavlova-9 [17]
Both A and C is true. However, even though a rectangle can be a parallelogram, a parallelogram is a quadrilateral, so the answer is A.
I hope this helps!
8 0
4 years ago
If the sides of a triangle are 3, 4, and 5, then, to the nearest degree, the measure of the smallest angle of the triangle is...
Tju [1.3M]

Answer:

The smallest angle of the triangle is 37\°

Step-by-step explanation:

we know that

If the sides of a triangle are 3, 4, and 5, then

we have a right triangle

because, applying the Pythagoras theorem

5^{2}=3^{2}+4^{2}

25=25 -------> is true

Remember that the measure of the smallest angle of triangle is the opposite  angle to the smallest side

so

Let

A-------> the measure of the smallest angle

tan(A)=3/4

A=arctan(3/4)=36.87\°

Round to the nearest degree

A=37\°

8 0
3 years ago
Read 2 more answers
What is 5/9 divided by 7/9
N76 [4]

Answer:

5/7

Step-by-step explanation

When dividing a fraction by a fraction it is the same as multiply the inverse of the bottom fraction, so 5/9 times 9/7 so the 9's cancel and your left with 5/7

4 0
4 years ago
What are three rational numbers that have a value between 1 and 1.5?
gulaghasi [49]
The answer is 1.1, 1.3, and 1.13 can be used
7 0
4 years ago
Which equations represent the asymptotes of the hyperbola (x-1)^2/36-(y-2)^2/64=1 ?
allochka39001 [22]

Answer:

4x+3y=10 and  4x-3y=-2

Step-by-step explanation:

The asymptote equation of a translate hyperbola with equation \frac{(x-h)^2}{b^2}-\frac{(y-k)^2}{a^2}=1 is y=\pm\frac{b}{a}(x-h)+k

The given hyperbola has equation: \frac{(x-1)^2}{36}-\frac{(y-2)^2}{64}=1

Or \frac{(x-1)^2}{6^2}-\frac{(y-2)^2}{8^2}=1

Comparing to \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

We have  a=6 and b=8, h=1 and k=2.

We substitute this values to get:

y=\pm\frac{6}{8}(x-1)+2

y=\pm\frac{4}{3}(x-1)+2

3y=\pm4(x-1)+6

We split the plus or minus sign to get:

3y=4x-4+6

3y=4x+2

3y-4x=2

Or

3y=-4(x-1)+6

3y=-4x+4+6

3y+4x=10

5 0
4 years ago
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