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just olya [345]
2 years ago
10

How much would your description cost for entire year,including 13% tax

Mathematics
2 answers:
kkurt [141]2 years ago
8 0
I don’t think enough information is provided
Korolek [52]2 years ago
6 0

Answer:

Yes

Step-by-step explanation:

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Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through
Setler79 [48]

<u>ANSWER:  </u>

The area of the triangle bounded by the y-axis is  \frac{7938}{4225} \sqrt{65} \text { unit }^{2}

<u>SOLUTION:</u>

Given, f(x)=9-\frac{-4}{7} x

Consider f(x) = y. Hence we get

f(x)=9-\frac{-4}{7} x --- eqn 1

y=9-\frac{4}{7} x

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = \frac{1}{ab} where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = \frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}

So, slope of perpendicular line = \frac{-1}{\text {slope of } f(x)}=\frac{7}{4}

Perpendicular line is passing through origin(0,0).So by using point slope formula,

y-y_{1}=m\left(x-x_{1}\right)

Where m is the slope and \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)

y-0=\frac{7}{4}(x-0)

y=\frac{7}{4} x --- eqn 2

4y = 7x

7x – 4y = 0  

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}

Put x value in (2)

\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}

So, the point of intersection is \left(\frac{252}{65}, \frac{441}{65}\right)

Length of f(x) is distance between \left(\frac{252}{65}, \frac{441}{65}\right) and (0,9)

\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}

Now, length of perpendicular of f(x) is distance between \left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)

\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}

Now, area of right angle triangle = \frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}

=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}

Hence, the area of the triangle is \frac{7938}{4225} \sqrt{65} \text { unit }^{2}

8 0
3 years ago
The sum of two expressions
vagabundo [1.1K]

Answer:

4x + 2y

Step-by-step explanation:

If you subtract 5x + 3y from 9x + 5y, you get:

  9x + 5y

<u> - 5x + 3y</u>

  4x + 2y

7 0
2 years ago
What is the cosine of 45°
Degger [83]
Cosine 45° = √2/2 (Square root of 2 over 2)
4 0
3 years ago
Decide if the following is a true or false statement - explain how you know
Sophie [7]

Answer:

False

Step-by-step explanation:

The equation equals -1/2 - 4.25 which is -4.75, which is negative.

Hope that helps!

5 0
3 years ago
NO LINKS!!! Please help fill in the blanks. Part 11a
Lorico [155]

Answer:

  see attached

Step-by-step explanation:

Here's your worksheet with the blanks filled.

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Of course, you know these log relations:

  log(a^b) = b·log(a) . . . . . power property

  log(a/b) = log(a) -log(b) . . . . . quotient property

  log(x) = log(y)   ⇔   x = y . . . . . . . . . equality property

4 0
2 years ago
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