All categories

3 years ago

Use one or more properties to rewrite each expression as an expression that does not use parentheses.

Mathematics

1 answer:

maria [59]3 years ago

4 0

Answer:

b + 9

Step-by-step explanation:

Given:

(b + 3) + 6

Required:

Rewrite expression without the parentheses.

SOLUTION:

(b + 3) + 6

Eliminate the parentheses

b + 3 + 6

Add like terms

b + 9

Thus:

(b + 3) + 6 = b + 9

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The arch of a bridge forms the upper half of an ellipse. The arch is 6 meters above the 20-meter-wide river. Write an equation f

aksik [14]

Answer:

$\frac{x^2}{36} + \frac{y^2}{100} = 1$

Step-by-step explanation:

Required

The equation of the ellipse

This is calculated as:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Where:

$b =6$ ---- height

$a = 0.5 * 20 =10$ --- width

So:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\frac{x^2}{6^2} + \frac{y^2}{10^2} = 1$

$\frac{x^2}{36} + \frac{y^2}{100} = 1$

3 0

3 years ago

How to write 12.375 in expanded form

posledela

Twelve and three hundred and seventy five thousandths.

3 0

3 years ago

Read 2 more answers

Plz help: What is 275 time 9000.

Kamila [148]

2,475,000 this your answer

8 0

3 years ago

Read 2 more answers

Is this correct? If not, please tell me the correct answer!!!

White raven [17]

Answer:

That is correct

Step-by-step explanation:

180° rotation about origin just means that you have to flip the pre-existing image across the x-axis

Good job!

3 0

3 years ago

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