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3 years ago

I NEED HELP ASAP CAUSE I'M STUCK

Mathematics

1 answer:

MaRussiya [10]3 years ago

4 0

Answer:

5. $4b^8c^3$

6. $30t^6v^1^0$

Step-by-step explanation:

5. $(-4b^6)(-b^2c^3)\\4b^8c^3$

6. $(10t^4v^5)(3t^2v^5)\\30t^6v^1^0$

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antiseptic1488 [7]

Do you have the rest of the question?

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3 years ago

Solve the system of linear equations using the Gauss-Jordan elimination method.

mixas84 [53]

Answer:

x = -0.6

y = 2.2

z = 2

Step-by-step explanation:

2x + y - 2z = -3

x + 3y - z = 4

3x + 4y - z = 5

Rewrite the system in matrix form and solve it by Gaussian Elimination (Gauss-Jordan elimination)

2 1 -2 -3

1 3 -1 4

3 4 -1 5

R1 / 2 → R1 (divide the 1 row by 2)

1 0.5 -1 -1.5

1 3 -1 4

3 4 -1 5

R2 - 1 R1 → R2 (multiply 1 row by 1 and subtract it from 2 row); R3 - 3 R1 → R3 (multiply 1 row by 3 and subtract it from 3 row)

1 0.5 -1 -1.5

0 2.5 0 5.5

0 2.5 2 9.5

R2 / 2.5 → R2 (divide the 2 row by 2.5)

1 0.5 -1 -1.5

0 1 0 2.2

0 2.5 2 9.5

R1 - 0.5 R2 → R1 (multiply 2 row by 0.5 and subtract it from 1 row); R3 - 2.5 R2 → R3 (multiply 2 row by 2.5 and subtract it from 3 row)

1 0 -1 -2.6

0 1 0 2.2

0 0 2 4

R3 / 2 → R3 (divide the 3 row by 2)

1 0 -1 -2.6

0 1 0 2.2

0 0 1 2

R1 + 1 R3 → R1 (multiply 3 row by 1 and add it to 1 row)

1 0 0 -0.6

0 1 0 2.2

0 0 1 2

x = -0.6

y = 2.2

z = 2

7 0

3 years ago

In f(x) = 2x2 − 8x − 10, the y-intercept is ? at ? and the x-intercepts are (-1, 0) and ? .

IgorLugansk [536]

<h2>1. y-intercept</h2>

$\boxed{(0,-10)}$

The quadratic function $f(x)=2x^22-8x-10$ represents a parabola. In fact, the graph of a quadratic function is a special type of U-shaped curve called a parabola. To find the y intercept, we set $x=0$ as follows:

$f(x)=2x^2-8x-10 \\ \\ If \ x=0 \rightarrow f(0)=-10 \\ \\ Then \ y-intercept: \\ \\ (0,-10)$

<h2>2. x-intercepts</h2>

$\boxed{(-1,0) \ and \ (5,0)}$

To find the other x-intercept, we must set $y=0$ as follows:

$f(x)=2x^2-8x-10 \\ \\ If \ y=0 \rightarrow 2x^2-8x-10=0 \\ \\ Using \ the \ quadratic \ formula: \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \therefore x=\frac{-(-8) \pm \sqrt{(-b)^2-4(2)(-10)}}{2(2)} \\ \\ \therefore x_{1}=-1 \ and \ x_{2}=5$