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3 years ago

The sum of two numbers is 59 . The smaller number is 17 less than the larger number.

Mathematics

1 answer:

VikaD [51]3 years ago

5 0

N+u=59
u+17=n
38=n
21=u
38+21=59

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Read 2 more answers

Angles α and β are angles in standard position such that: α terminates in Quadrant I and sinα = 3/5 β terminates in Quadrant III

Arte-miy333 [17]

Answer:

-16/65

Step-by-step explanation:

Given sinα = 3/5 in quadrant 1;

Since sinα = opp/hyp

opp = 3

hyp = 5

adj^2 = hyp^2 - opp^2

adj^2 = 5^2 = 3^2

adj^2 = 25-9

adj^2 = 16

adj = 4

Since all the trig identity are positive in Quadrant 1, hence;

cosα = adj/hyp = 4/5

Similarly, if tanβ = 5/12 in Quadrant III,

According to trig identity

tan theta = opp/adj

opp = 5

adj = 12

hyp^2 = opp^2+adj^2

hyp^2 = 5^2+12^2

hyp^2 = 25+144

hyp^2 = 169

hyp = 13

Since only tan is positive in Quadrant III, then;

sinβ = -5/13

cosβ = -12/13

Get the required expression;

sin(α - β) = sinαcosβ - cosαsinβ

Substitute the given values

sin(α - β) = 3/5(-12/13) - 4/5(-5/13)

sin(α - β)= -36/65 + 20/65

sin(α - β) = -16/65

Hence the value of sin(α - β) is -16/65

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3 years ago

Answer Now!! Which expression can be used to find the area, in square cm, of this parallelogram?

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3 years ago

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