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Salsk061 [2.6K]

3 years ago

6

The appearance of text is called hyperlink. True False

1 answer:

Lerok [7]3 years ago

8 0

The answer is True hope this helps

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Whats 125% as a fraction and a decimal

ZanzabumX [31]

125% as a fraction is 1 25/100 and in decimal form it is 1.25.

5 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

What is the reciprocal of 8 8/5 ?

Ksenya-84 [330]

Answer:

$\frac{5}{48}$

Step-by-step explanation:

The reciprocal of a number n is $\frac{1}{n}$

Given 8 $\frac{8}{5}$ , change to an improper fraction

= $\frac{5(8)+8}{5}$ = $\frac{48}{5}$ , thus

reciprocal = $\frac{1}{\frac{48}{5} }$ = $\frac{5}{48}$

3 0

3 years ago

If the total value of the coins is 36¢, what is the missing coin?

laiz [17]

Answer:

1 penny 3 nickels 2 dimes = 36 cents

quarter

Step-by-step explanation:

8 0

2 years ago

Find an equation of the plane orthogonal to the line

jolli1 [7]

The given line is orthogonal to the plane you want to find, so the tangent vector of this line can be used as the normal vector for the plane.

The tangent vector for the line is

d/d<em>t</em> (⟨0, 9, 6⟩ + ⟨7, -7, -6⟩<em>t </em>) = ⟨7, -7, -6⟩

Then the plane that passes through the origin with this as its normal vector has equation

⟨<em>x</em>, <em>y</em>, <em>z</em>⟩ • ⟨7, -7, -6⟩ = 0

We want the plane to pass through the point (9, 6, 0), so we just translate every vector pointing to the plane itself by adding ⟨9, 6, 0⟩,

(⟨<em>x</em>, <em>y</em>, <em>z</em>⟩ - ⟨9, 6, 0⟩) • ⟨7, -7, -6⟩ = 0

Simplifying this expression and writing it standard form gives

⟨<em>x</em> - 9, <em>y</em> - 6, <em>z</em>⟩ • ⟨7, -7, -6⟩ = 0

7 (<em>x</em> - 9) - 7 (<em>y</em> - 6) - 6<em>z</em> = 0

7<em>x</em> - 63 - 7<em>y</em> + 42 - 6<em>z</em> = 0

7<em>x</em> - 7<em>y</em> - 6<em>z</em> = 21

so that

<em>a</em> = 7, <em>b</em> = -7, <em>c</em> = -6, and <em>d</em> = 21

4 0

3 years ago