Ask question

Ask question

All categories

3 years ago

7

Teresa is factoring this polynomial by grouping. Which common factors should be used in the next step of factoring? 10x3 3x2−20x

−6 (10x3 3x2) (−20x−6) x2 and −2x 2x2 and −2x x2 and −2 2x2 and −2.

2 answers:

Oksi-84 [34.3K]3 years ago

8 0

Answer:

−2=−2x^3

Step-by-step explanation:

10x3= 3x2−20x−6=(10x3⋅(3x2))(−20x−6)=−600x6−180x5x2=−600x6−180x5−2x⋅(2x2)=−4x3−2x⋅x2=−2x^3

−2=−2x3

2x^2=−2x^3

−2=−2x^3

Elanso [62]3 years ago

4 0

Answer:

> The correct choice is C) x2 and −2

Step-by-step explanation:

> This is the only answer choice that makes sense in this equation.

You might be interested in

Svetllana [295]

Answer:

The sale price was 60% of the purchase price.

Step-by-step explanation:

Given that Blake bought a motorcycle for $ 550 last year and sold it for $ 330 this year, to determine what is his sale price as a percentage of his purchase price, the following calculation must be performed:

550 = 100

330 = X

330 x 100/550 = X

33000/550 = X

60 = X

Therefore, the sale price was 60% of the purchase price.

5 0

3 years ago

Please help me with this question

MA_775_DIABLO [31]

Answer:

B. 36

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Helpppppppppppppppppp pls

BartSMP [9]

Answer:

b=30

Step-by-step explanation:

The angle here is a right angle which is an angle that equals 90 degrees. You subtract that measure of the angle by 60 and you end up with 30.

7 0

3 years ago

Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

5 0

3 years ago

I dont really know how to do soh cah toah. please help me find x.

Ipatiy [6.2K]

First notice that the triangle with sides $x,y,a$ and the triangle with sides $z,a+b,x$ are similar. This is true because the angle between sides $x,y$ in the smaller triangle is clearly $30^\circ$ , while the angle between sides $y,z$ in the larger triangle is clearly $60^\circ$ . So the triangles are similar with sides $x,y,a$ corresponding to $a+b,z,x$ , respectively.

Now both triangles are $30^\circ-60^\circ-90^\circ$ , which means there's a convenient ratio between its sides. If the length of the shortest leg is $\ell_1$ , then the length of the longer leg is $\ell_2=\sqrt3\ell_1$ and the hypotenuse has length $\ell_3=\sqrt{{\ell_1}^2+{\ell_2}^2}=2\ell_1$ .

Since $x$ is the shortest leg in the larger triangle, it follows that $a+b=2x$ , so $a+b=15=2x\implies x=\dfrac{15}2$

4 0

4 years ago