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3 years ago

In △ABC,c=71, m∠B=123°, and a=65. Find b. A. 101.5 B. 117.8 C. 123.0 D. 119.6

Mathematics

1 answer:

tia_tia [17]3 years ago

8 0

Answer:

Option D

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other the given angle is the included angle. This means that the angle B is formed by the intersection of the lines a and c. Therefore, cosine rule will be used to calculate the length of b. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that c=71, B=123°, and a=65. Plugging in the values:

b^2 = 65^2 + 71^2 - 2(65)(71)*cos(123°).

Simplifying gives:

b^2 = 14293.0182932.

Taking square root on the both sides gives b = 119.6 (rounded to the one decimal place).

This means that the Option D is the correct choice!!!

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First you to find the worksheet and download it<br> plase I need help

Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = $3^{x + 2}$

The asymptote is given as x → -∞, f(x) = $3^{x + 2}$ → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = $3^{0 + 2}$ = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = $5^{1 - x}$

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = $5^{1 - 0}$ = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = $2^{x + 3}$ - 1

The asymptote is $2^{x + 3}$ → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) = $2^{0 + 3}$ - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, $2^{x + 3}$ - 1 = 0

$2^{x + 3}$ = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) $f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2$

The asymptote is $\left (\dfrac{1}{2} \right)^{x - 2}$ → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = $f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6$

The y-intercept is (0, 6)

7 0

2 years ago

For the following exercises, find (f ∘ g)(x) and (g ∘ f)(x) for each pair of functions.

Rufina [12.5K]

Answer:

The value of $(f \circ g)(x)$ is 17-18x and $(g \circ f)(x)$ is -7-18x.

Step-by-step explanation:

It is given in the question functions f(x) as 3x+2 and g(x)=5-6x.

It is required to find $(f \circ g)(x)$ and $(g \circ f)(x)$ .

To find $(f \circ g)(x)$ , substitute g(x) for x in f(x) and simplify the expression.

To find $(g \circ f)(x)$ , substitute f(x) for x in g(x) and simplify the expression.

Step 1 of 2

Substitute g(x) for x in f(x) and simplify the expression.

$\begin{aligned}&(f \circ g)(x)=f(5-6 x) \\&(f \circ g)(x)=3(5-6 x)+2 \\&(f \circ g)(x)=15-18 x+2 \\&(f \circ g)(x)=17-18 x\end{aligned}$

Step 2 of 2

Substitute f(x) for x in g(x) and simplify the expression.

$\begin{aligned}&(g \circ f)(x)=g(3 x+2) \\&(g \circ f)(x)=5-6(3 x+2) \\&(g \circ f)(x)=5-18 x-12 \\&(g \circ f)(x)=-7-18 x\end{aligned}$

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2 years ago

What is the product?

ZanzabumX [31]

Answer:

The product is:

$\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]$

Step-by-step explanation:

For this problem you need to multiply the first row only for the two first column of the others matrix and get the desired result:

$\left[\begin{array}{ccc}1&3&1\\-2&1&0\end{array}\right] \times \left[\begin{array}{cc}2&-2\\3&5\\4&1\end{array}\right]$

$1 \times 2 + 3 \times 3 + 1 \times -2 = 15$

So the value of the element in the position $a_{11}$ is 15

$1 \times -2 + 3 \times 5 + 1 \times 1 = 14$

So the value of the element in the position $a_{12}$ is 14

Then with these two values you can determinate the result matrix.

$\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]$

3 0

2 years ago

(10-P) Help please, thanks.

Anuta_ua [19.1K]

The formula of linear equation is:
$y1 - y2 = m(x1 - x2)$
Where
x1 and x2: x coordinates(-1 and 3)
y1 and y2: y coordinates(-2 and 10)
m is the slope

We can then choose a point from the line to find the eqaution.

We need to first find the slope:
$\frac{y1 - y2}{x1 - x2}$
In this case:
$\frac{ - 2 - 10}{ - 1 - 3} \\ = \frac{ - 12}{ - 4} \\ = 3$

In this case, as the y2 is given as 1,put (-1,-2) , and the slope(3 )into the eqaution :

y-(-2) = 3(x-(-1))
y+2 = 3(x+1)

Therefore
$y + 2 = 3x + 1$
is the answer.

Hope it helps!

7 0

3 years ago

G(n)=-4n+2; find g(8)<br>

yarga [219]

Answer:

-30

Step-by-step explanation:

In g(8) the 8 = n. So, substitute 8 in for n in the expression -4n + 2

g(8) = -4(8) + 2 = -32 + 2 = -30

4 0

3 years ago

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