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2 years ago

Ok let try this plz ....

Mathematics

1 answer:

satela [25.4K]2 years ago

6 0

Answer:

Should be A

Step-by-step explanation:

It would be A because you would make the original x coordinates negative to get the new ones

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Which of the following is not true?

Lisa [10]

Answer:

The last one.

Step-by-step explanation: Because 1/2 divided by 4 = 1/8 and 1/2 x 4/1 = 2

3 0

3 years ago

Pls answer precisely

Nata [24]

Answer:

The third one

Step-by-step explanation:

Its asking if its positive in the interval [-3, -2]

Which means you only look at the graph from -3, -2 on the x axis and we are looking see which one is positive in that space. And by positive it means going up in a positive direction. Only the third one does this.

3 0

2 years ago

In order to pay off he car she bought lauri had to make 34 more payments of $145.98. How much does she still owe?

Talja [164]

4963.32 is the answer your looking for man.

8 0

3 years ago

Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0

Alik [6]

Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

$(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0$

The general equation for the second-order polynomial is:

$x = \frac{2\cdot k \pm \sqrt{4\cdot k^{2}-4\cdot (k^{2}-1)\cdot (-3\cdot k + 1)}}{k^{2}-1}$

This condition must be observed for the case of a quadratic equation with equal roots:

$4\cdot k^{2} - 4\cdot (k^{2}-1)\cdot (-3\cdot k + 1) = 0$

$k^{2} + (k^{2}-1)\cdot (3\cdot k + 1) = 0$

$k^{2} + 3\cdot k^{3} - 3\cdot k - k^{2}-1 = 0$

$3\cdot k^{3} - 3\cdot k - 1 = 0$

7 0

3 years ago

How do you solve 8a squared plus 2 = 634

Inga [223]

Answer:

a ≈ 8.9

Step-by-step explanation:

Set up the equation as so:

8a^2 + 2 = 634

First, subtract two from both sides:

8a^2 = 632

Then, divide by 8 to further isolate the variable.

a^2 = 79

To get rid of the squared, you have to take the square root of both sides. The square root of 79 is roughly 8.9. Ergo, a ≈ 8.9

8 0

3 years ago