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4 years ago

Please help! (ignore the answer I chose)

Mathematics

1 answer:

Andrei [34K]4 years ago

5 0

The numerator factors as (t-8)(t+4), so the whole thing is (t-8)(t+4)/(t-8). Now we can cancel the t-8 AS LONG as t isn't equal to 8, otherwise, we are canceling a zero from both numerator and denominator which is invalid. What is left is t+4.

So your choice was right.

You might be interested in

The total surface are of a can is 296.73in ^2. if the radius of the can is 4.5 inches what is the height of the can

Travka [436]

Surface Area of can (SA) = (2 · π · r²) + (2 · π · r · h)

296.73 = [2 · π · (4.5)²] + [2 · π · (4.5) · h]

296.73 = 40.5π + 9π · h

296.73 - 40.5π = 9π · h

(296.73 - 40.5π)/9π = h

169.5/28.27 = h

6 = h

Answer: 6 inches

5 0

3 years ago

What is the sum of the solutions of x2 + 9x + 20 = 0?

ZanzabumX [31]

Answer:

Sum of the solutions of $x^2+9x+20=0$ is -9.

Product of the solutions of $6x^2+7x=3$ is $-0.50$

Step-by-step explanation:

1. $x^2+9x+20=0$

Given:

The expression whose sum of the solution is required is given as:

$x^{2} +9x+20=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the sum of the solutions is given as:

Sum = $\frac{-b}{a}$

Here, $a=1,b=9,c=20$

Therefore, the sum of the solutions = $-\frac{9}{1}=-9$

2. $6x^2+7x=3$

Rewriting the above equation in a standard quadratic equation, we get:

$6x^2+7x-3=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the product of the solutions is given as:

Product = $\frac{c}{a}$

Here, $a=6,b=7,c=-3$

Therefore, the product of the solutions = $\frac{-3}{6}=-0.50$

6 0

3 years ago

Can someone help me with this 1st question please

lianna [129]

Answer:

Add 3 to X to get Y

Step-by-step explanation:

1 + 3 = 4

2 + 3 = 5

4 0

2 years ago

Read 2 more answers

Solve by completing the square

erastovalidia [21]

Answer:

Your answers would be:

x = 10 and x = -8

Step-by-step explanation:

Solve for the zeros:

x² - 2x - 78 = 2

Subtract both sides by 2.

x² - 2x - 80 = 0

Use the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = 1

b = -2

c = -80

Plug in and solve:

$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-80)}}{2(1)}\\\\x=\frac{2\pm\sqrt{(-2)^2-4(1)(-80)}}{2(1)}\\\\x=\frac{2\pm\sqrt{4-4(-80)}}{2(1)}\\\\x=\frac{2\pm\sqrt{4+320}}{2(1)}\\\\x=\frac{2\pm\sqrt{324}}{2(1)}\\\\x=\frac{2\pm18}{2}\\$

Solve for +:

$x=\frac{2+18}{2}\\\\x=\frac{20}{2}\\\\x=10$

Solve for -:

$x=\frac{2-18}{2}\\\\x=\frac{-16}{2}\\\\x=-8$

5 0

3 years ago

Which of the following represents the series ?

aleksklad [387]

The denominator in each term of the series is a square number, but neither $n-6$ nor $n^2+1$ are squares. So none of (1), (2), and (4) can be correct. The answer must be (3).

5 0

3 years ago