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3 years ago

What is the last thing you do as a check inside the car?

Mathematics

1 answer:

Vitek1552 [10]3 years ago

7 0

I always make sure my hand brake is on

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Gerry wants to construct a triangle with segments that

alexira [117]

Answer:

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Step-by-step explanation:

Sample Response: No, the triangle is not possible. The triangle inequality rule states the sum of any two sides must be greater than the third side. 6 + 4 is 10, which is less than 11.

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2 years ago

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2 years ago

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postnew [5]

Answer:

50°

Step-by-step explanation:

Hope this helps and is correct

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2 years ago

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Need help with school

Scorpion4ik [409]

Answer:

x = 7.2m

Step-by-step explanation:

Because it is a right triangle, you can use the Pythagorean theorem:

4^2+6^2 = x^2

16+36 = x^2

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2 years ago

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Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago