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podryga [215]
3 years ago
12

Why is this not in quadratic form? x^6+ 5x^4 + 6

Mathematics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer and Step-by-step explanation:

Quadratic form has only the exponent of 2 and 1 in it. An example of it is as shown:

3x^2 + 4x - 12

Your expression isn't a quadratic since it has a 6 as the highest exponent, but it is a trinomial (an expression that has 3 terms).

<u><em>#teamtrees #PAW (Plant And Water)</em></u>

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Find an expression for the area enclosed by quadrilateral ABCD below.
viktelen [127]

∆ABD is right angled hence area:-

\\ \sf\longmapsto \dfrac{1}{2}bh

\\ \sf\longmapsto \dfrac{1}{2}(4x)(3x)

\\ \sf\longmapsto \dfrac{1}{2}(12x^2)

\\ \sf\longmapsto 6x^2

There is only one option containing 6x^2 i.e Option D.

Hence without calculating further

Option D is correct

8 0
2 years ago
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Can someone help me with this geometry question asap, thanks!
ivann1987 [24]

Answer:

D

Step-by-step explanation:

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6 0
3 years ago
Please help me with this problem......
kupik [55]

The red graph is the result of a vertical translation, 4 units down, of the black graph.

This means that, whenever the black graph associates

x\mapsto f(x)

The red graph must associate 4 less to the same input:

x\mapsto f(x)-4

So, the equation for the red graph is

y=f(x)-4

or, equivalently,

y+4=f(x)

7 0
3 years ago
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Find the 17th term in the sequence for which a1 = 10 and d = -3.
yKpoI14uk [10]

Answer:

d. -38

Step-by-step explanation:

Arithmetic sequence concepts:

The general rule of an arithmetic sequence is as follows:

a_{n+1} = a_{n} + d

In which d is the common diference between each term.

We can expand the general equation to find the nth term from the first, by the following equation:

a_{n} = a_{1} + (n-1)*d

In this question:

a_{1} = 10, d = -3

17th term:

a_{17} = a_{1} + (17-1)*d = 10 + 16*(-3) = -38

So the correct answer is:

d. -38

6 0
3 years ago
How do I find the restrictions on x if there are any? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bx%20-%201%7D%20%20%3
Natali5045456 [20]

We have the expression:

\frac{1}{x - 1}=\frac{5}{x - 10}

When we have rational functions, where the denominator is a function of x, we have a restriction for the domain for any value of x that makes the denominator equal to 0.

That is because if the denominator is 0, then we have a function f(x) that is a division by zero and is undefined.

If we have a value that makes f(x) to be undefined, then this value of x does not belong to the domain of f(x).

Expression:

\begin{gathered} \frac{1}{x-1}=\frac{5}{x-10} \\ \frac{x-1}{1}=\frac{x-10}{5} \\ x-1=\frac{x}{5}-\frac{10}{5} \\ x-1=\frac{1}{5}x-2 \\ x-\frac{1}{5}x=-2+1 \\ \frac{4}{5}x=-1 \\ x=-1\cdot\frac{5}{4} \\ x=-\frac{5}{4} \end{gathered}

Answer: There is no restriction for x in the expression.

6 0
1 year ago
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