Factor this polynomial expression.

Which one is correct

Nitella [24]

Answer:

Quadrilateral

Step-by-step explanation:

The question is straightforward, and it requires a direct answer.

A rhombus has four sides and as such it means it is a type of quadrilateral.

So, before a shape is proven to be a rhombus, it has to first be proven to be a quadrilateral.

4 0

3 years ago

4. Solve the equation using the quadratic formula.

Vikentia [17]

Answer:

A. x = -2, x = 1.25

Step-by-step explanation:

Use the quadratic formula

x = $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

4x² + 3x - 10 = 0

a = 4

b = 3

c = - 10

x = $\frac{-3+\sqrt{3^{2}-4x4(-10) } }{2x4}$

Simplify

Evaluate the exponent

x = $\frac{-3+\sqrt{9-4x4(-10)} }{2x4}$

Multiply the numbers

x = $\frac{-3+\sqrt{9+160} }{2x4}$

Add the numbers

x = $\frac{-3+\sqrt{169} }{2x4}$

Evaluate the square root

x = $\frac{-3+13}{2x4}$

Multiply the numbers

x = $\frac{-3+13}{8}$

Separate the equations

To solve for the unknown variable, separate into two equations: one with a plus and the other with a minus.

x = $\frac{-3+13}{8}$

x = $\frac{-3-13}{8}$

Solve

Rearrange and isolate the variable to find each solution

x = - 2

x = 1.25

6 0

3 years ago

Read 2 more answers

If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1 help meeeee plessss

Trava [24]

$A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}$

where det(A) = 1×1 - 2×1 = -1.

$B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}$

where det(B) = 0×2 - (-1)×1 = 1. Then

$B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}$

On the other side, we have

$AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}$

and det(AB) = det(A) det(B) = (-1)×1 = -1. So

$(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}$

and both matrices are clearly the same.

More generally, we have by definition of inverse,

$(AB)(AB)^{-1} = I$

where $I$ is the identity matrix. Multiply on the left by A ⁻¹ to get

$A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}$

Multiplication of matrices is associative, so we can regroup terms as

$(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}$

Now multiply again on the left by B ⁻¹ and do the same thing:

$B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}$

7 0

3 years ago

Which angles are supplementary to 14 ? select all that apply.

Furkat [3]

Answer will be 15>
Supplementary angles are two angles with the sun of 90 degrees

5 0

3 years ago

Read 2 more answers

I need answers please.

Stells [14]

Problem 1

a = length of side 1
b = length of side 2
c = length of side 3

Let's say that c = 36.

Since she has 90 feet of rope total, that leaves 90-36 = 54 feet of rope for sides 'a' and b to divide up somehow. We can say a+b = 54

Now let's say we want 'a' to be as large as possible. To do this, we need to make b as small as possible. That occurs when b = 1. We can't have b = 0, or else a triangle won't form. So the next value up is b = 1.

If b = 1, then,

a+b = 54

a = 54-b

a = 54-1

a = 53

The longest side possible is 53 feet

<h3>Answer: 53 feet</h3>

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Problem 2

Recall that (5,12,13) is a pythagorean triple. This is because 5^2+12^2 = 13^2. Both sides lead to 169.

Since we have a pythagorean triple, this indicates a triangle with sides 5,12,13 is a right triangle.

If a triangle has sides 5,12, and 13, then the perimeter is 5+12+13 = 30.

Triple each side to get a larger triangle of 15, 36, and 39. The larger perimeter is 90 after adding those larger sides. Note the jump from 30 to 90 is "times 3". It's not a coincidence that the perimeter has multiplied by the same scale factor as the side lengths.

Also note that 15^2+36^2 = 39^2. Both sides lead to 1521. This indicates we have a right triangle and (15,36,39) is another pythagorean triple. Any scaled version of a pythagorean triple, is also a pythagorean triple.

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In short, we've found that a triangle with sides 15, 36, and 39 is a right triangle and has perimeter 90 feet.

<h3>Answer: 15 ft, 36 ft, 39 ft</h3>

8 0

3 years ago