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aalyn [17]
3 years ago
9

5364982+574=82983675

Mathematics
1 answer:
UNO [17]3 years ago
5 0
Yes. Yes yesyesyes. Yes. Very yes
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Which is the graph of a quadratic equation that has a negative discriminant?
Natasha2012 [34]

Answer:

D

Step-by-step explanation:

A negative discriminant indicates that the quadratic equation has no real roots.

Thus the graph does not touch or intersect the x- axis

The only graph that does not touch or intersect the x- axis is the fourth one

5 0
3 years ago
Read 2 more answers
I don't get this pls help
dmitriy555 [2]

Answer:

  d, e

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  a^-b = 1/a^b

__

In this case, it means the product is ...

  (6^1)(6^0)(6^-3) = 6^(1+0-3) = 6^-2 = 1/6^2 = 1/36

__

The 6 without an exponent is equivalent to 6^1, an exponent of 1.

The sum of the exponents is -2.

Add the exponents to simplify the expression.

The value of the expression is 1/36.

An equivalent is any expression that results in 6^-2. One such is (6^5)(6^-7).

__

Only the last two choices, d and e, apply.

3 0
3 years ago
The terminal side of an angle in standard position passes through p(–3, –4). what is the value of tangent theta?
astraxan [27]

Answer:

  4/3

Step-by-step explanation:

The tangent of any angle (θ) in standard position that has point (x, y) on its terminal ray is ...

  tan(θ) = y/x

__

For the given point on the terminal side, the tangent is ...

  tan(θ) = (-4)/(-3) = 4/3

_____

<em>Additional comment</em>

There are several ways this can be explained. One of them makes use of the relation between rectangular and polar coordinates:

  (x, y) = (r·cos(θ), r·sin(θ))

Then the ratio y/x is ...

  y/x = (r·sin(θ))/(r·cos(θ)) = sin(θ)/cos(θ) = tan(θ)

6 0
1 year ago
Read 2 more answers
Given that f (x)=2x+5 and g(x)=x-7 solve for f (g (x)) when x=-3
Alexeev081 [22]
2(-3-7)+5= <span>2(-10)+5= -20 + 5= -15

-15 is your final answer</span>
7 0
3 years ago
4(3x^2y^4)^3 / (2x^3y^5)^4
tatyana61 [14]

Solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Step-by-step explanation:

We need to Solve the expression: \frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

Solving:

\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}

Applying exponent rule: \frac{x^a}{x^b}=x^{a-b}

=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}

Another exponent rule says: x^{-a}=\frac{1}{x^a}=

=\frac{27}{4x^{6}y^{8}}

So, solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Keywords: Solving Exponents  

Learn more about Solving Exponents at:

  • brainly.com/question/13174260
  • brainly.com/question/13174254
  • brainly.com/question/13174259

#learnwithBrainly

8 0
3 years ago
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