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3 years ago

What is 3/4 foot long x0.5

Mathematics

2 answers:

Aleks [24]3 years ago

8 0

I think it is 11 thats what i think

vivado [14]3 years ago

4 0

Well the problem should be 3/4 x 0.5 so we need o convert 3/4 to 0.75 then multiply that by 0.5 which would be .375

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How do you answer this MATH

raketka [301]

X = 5 or x = 13.

If you need details, please ask!

7 0

3 years ago

Read 2 more answers

Find the value of sin pi/4-x cos pi/4-y + cospi/4-x sin pi/4-y

defon

Annaba jajsjdhs jakka dusk

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2 years ago

Lines k and n intersect on the y-axis

avanturin [10]

a) The equation of line k is:

$y = -\frac{202}{167}x + \frac{598}{167}$

b) The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

The equation of a line, in <u>slope-intercept formula</u>, is given by:

$y = mx + b$

In which:

m is the slope, which is the rate of change.
b is the y-intercept, which is the value of y when x = 0.

Item a:

Line k intersects line m with an angle of 109º, thus:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

In which $m_1$ and $m_2$ are the slopes of <u>k and m.</u>

Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

$m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}$

The tangent of 109 degrees is $\tan{109^{\circ}} = -\frac{29}{10}$
Thus, the slope of line m is found solving the following equation:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

$-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}$

$m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2$

$m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}$

$\frac{167m_2}{80} = \frac{-202}{80}$

$m_2 = -\frac{202}{167}$

Thus:

$y = -\frac{202}{167}x + b$

It goes through point (-2,6), that is, when $x = -2, y = 6$ , and this is used to find b.

$y = -\frac{202}{167}x + b$

$6 = -\frac{202}{167}(-2) + b$

$b = 6 - \frac{404}{167}$

$b = \frac{6(167)-404}{167}$

$b = \frac{598}{167}$

Thus. the equation of line k, in slope-intercept formula, is:

$y = -\frac{202}{167}x + \frac{598}{167}$

Item b:

Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

$-\frac{202}{167}m = -1$

$m = \frac{167}{202}$

Then

$y = \frac{167}{202}x + b$

Also goes through point (-2,6), thus:

$6 = \frac{167}{202}(-2) + b$

$b = \frac{(2)167 + 202(6)}{202}$

$b = \frac{1546}{202}$

The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

A similar problem is given at brainly.com/question/16302622

7 0

2 years ago

Graph XY with endpoints X(5,−2) and Y(3,−3) and its image after a reflection in the x-axis and then a rotation of 270 degrees ab

DerKrebs [107]

Answer:

X''(2, -5), Y''(3, -3)

Step-by-step explanation:

You know that reflection in the x-axis changes the sign of the y-coordinate. Points that used to be above the axis are now below by the same amount, and vice versa.

Rotation counterclockwise by 270° is the same as clockwise rotation by 90°. That maps the coordinates like this:

(x, y) ⇒ (y, -x)

The two transformations together give you ...

(x, y) ⇒ (x, -y) ⇒ (-y, -x) . . . . . . . . equivalent to reflection across y=-x.

Using this mapping, we have ...

X(5, -2) ⇒ X''(2, -5)

Y(3, -3) ⇒ Y''(3, -3) . . . . . . on the equivalent line of reflection, so invariant

_____

The attachment shows the original segment in red, the reflected segment in purple, and the rotated segment in blue. The equivalent line of reflection is shown as a dashed green line.

8 0

3 years ago

DIscrete Math

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

3 0

2 years ago