8 and 15 are the lengths of the legs.
The tangent is the ratio of the lengths of legs.
In general, tan x = opp/adj
You have 8/15, so here 8 is the opposite leg and 15 is the adjacent leg.
Angle A is the right angle, so that is not it.
For which acute angle, B or C, is 8 opposite and 15 adjacent?
Answer: for angle B.
Therefore, 8/15 must be tan B.

6 0

3 years ago

Read 2 more answers

Using the rule y = 4x - 1, find the value of y when x is 5.

blagie [28]

I think the answer is c 19

7 0

4 years ago

Read 2 more answers

Evaluate C_n.xP^xQn-x For the given n=7, x=2, p=1/2

r-ruslan [8.4K]

Answer:

The value of given expression is $\frac{21}{128}$ .

Step-by-step explanation:

Given information: n=7, x=2, p=1/2

$q=1-p=1-\frac{1}{2}=\frac{1}{2}$

The given expression is

$C(n,x)p^xq^{n-x}$

It can be written as

$^nC_xp^xq^{n-x}$

Substitute n=7, x=2, p=1/2 and q=1/2 in the above formula.

$^7C_2(\frac{1}{2})^2(\frac{1}{2})^{7-2}$

$\frac{7!}{2!(7-2)!}(\frac{1}{2})^2(\frac{1}{2})^{5}$

$\frac{7!}{2!5!}(\frac{1}{2})^{2+5}$

$\frac{7\times 6\times 5!}{2\times 5!}(\frac{1}{2})^{2+5}$

$21(\frac{1}{2})^{7}$

$\frac{21}{128}$

Therefore the value of given expression is $\frac{21}{128}$ .

7 0

3 years ago

The equation of the line through point (- 4, 0) and parallel to the y-axis is

riadik2000 [5.3K]

The gradient of y axis is zero. Since the line is parallel to y axis, its gradient is also zero. This
y=mx+c
Where m is the gradient;
y=(0)x+c
y=c
Replacing for y and x;
0=0(-4)+c
c=0

7 0

3 years ago