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Delvig [45]
4 years ago
15

The range of F(x) = 5.2% is all positive real numbers. O A. True O B. False

Mathematics
1 answer:
kirza4 [7]4 years ago
8 0

Answer:

If im not mistaken its true

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What is the slope indicated in the table below<br> x 4 6 8 10<br> y 7 14 21 28
xeze [42]

Answer:

3/2

Step-by-step explanation:

So for the x it increases by +2. For the y it increases by +7. We do rise over run which is y over x. So it will be 3/2.

3 0
3 years ago
Read 2 more answers
Which of the following is a good representation of a plane?
NikAS [45]
A table top would be a good representation of a plain because it is a flat surface, and isn't a 3D object, if that makes sense. A box wouldn't be a plane because it has 6 different sides but a table only has one side and one plane. hope this helps :)
8 0
3 years ago
NEED HELP ASAP 70 POINTS
Nataliya [291]

Answer:

1) Option A) is correct

The given rational exponent expression is not simplified correctly as a radical expression is

x^{\frac{7}{4}}=\sqrt[7]{x^4}

2)Option A) is correct

That is (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

Step-by-step explanation:

1) Given that x^{\frac{7}{4}}=(\sqrt{x^7})^\frac{1}{4}

x^{\frac{7}{4}}=\sqrt[4]{x^7} is the correct answer but in the given problem they gave the RHS as wrong.

Therefore the given rational exponent expression is not simplified correctly as a radical expression is

x^{\frac{7}{4}}=\sqrt[7]{x^4}

2)Given that the rational exponent expression is (729x^3y^{-18})^{\frac{1}{6}}

To find it as a radical expression:

(729x^3y^{-18})^{\frac{1}{6}}=(729)^{\frac{1}{6}}(x^3)^{\frac{1}{6}}((y^{-18})^{\frac{1}{6}})

=3(x^{\frac{3}{6}})(y^{\frac{-18}{6}})

=3(x^{\frac{1}{2}})(y^{-3})

=\frac{3\sqrt{x}}{y^3}

Therefore (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

Therefore Option A) is correct

That is (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

3 0
3 years ago
PLEASE HELP QWQ AsAp with these 4 questions
den301095 [7]

Answer:

Step-by-step explanation:

I can't believe I'm doing this for 5 points, but ok!

For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:

\left[\begin{array}{ccccc}7&4&6&7&4\\-4&8&9&-4&8\\1&8&7&1&8\end{array}\right]  and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):

(7*8*7)+(4*9*1)+(6*-4*8)-[(1*8*6)+(8*9*7)+(7*-4*4)] which gives us:

392 + 36 - 192 - [48 + 504 - 112] which simplifies to

236 - 440 which is -204

On to the second one:

\left[\begin{array}{ccccc}-8&-4&-1&-8&-4\\1&7&-3&1&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-8*7*9)+(-4*-3*8)+(-1*1*9)-[(8*7*-1)+(9*-3*-8)+(9*1*-4)] which gives us:

-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to

-417 - 124 which is -541, choice c.

Now for the third one:

\left[\begin{array}{ccccc}-2&-2&-5&-2&-2\\2&7&-3&2&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-2*7*9)+(-2*-3*8)+(-5*2*9)-[(8*7*-5)+(9*-3*-2)+(9*2*-2)] which gives us:

-126+48-90-[-280+54-36] which simplifies to

-168 - (-262) which is 94, choice c again.

Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.

\left[\begin{array}{cc}-4&-5&\\-6&-8\\\end{array}\right] \left[\begin{array}{c}x\\y\\\end{array}\right] = \left[\begin{array}{c}-5\\-2\\\end{array}\right] and I found the inverse of the 2 x 2 matrix on the left.

Find the inverse by:

* finding the determinate

* putting the determinate under a 1

* multiply that by the "mixed up matrix (you'll see...)

First things first, the determinate:

|A| = (-4*-8) - (-6*-5) which simplifies to

|A| = 32 - 30 so

|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:

\frac{1}{2}\left[\begin{array}{cc}-8&5\\6&-4\end{array}\right]  (to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:

\left[\begin{array}{cc}-4&\frac{5}{2}\\3&-2\\\end{array}\right] Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:

\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{cc}-4&\frac{5}{2} \\3&-2\end{array}\right] *\left[\begin{array}{c}-5&-2\\\end{array}\right] and that right side multiplies out to

x = 20 - 5 which is

x = 15 and

y = -15 + 4 which is

y = -11

(It works, I checked it)

7 0
3 years ago
Rewrite 1/10,000 to the power of ten
g100num [7]

Answer:

10^4

Step-by-step explanation:

3 0
2 years ago
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