All categories

3 years ago

WILL MARK BRAINLIEST! PLEASE HELP

1 answer:

4 0

Part 1) In triangle PQR, mP= 43, PQ=7.5 and PR= 8.4. What is mR to the nearest degree?

we know that
the law of cosines establishes that
c²=a²+b²-2*a*b*cos C

in this problem

c=QR
a=PQ=7.5
b=PR=8.4
C=angle P=43°
c²=a²+b²-2*a*b*cos C----> c²=7.5²+8.4²-2*7.5*8.4*cos 43°
c²=34.66--------> c=5.89
QR=5.9

applying the law of sines
7.5/sin R=5.9/sin 43-----> 7.5*sin 43=5.9*sin R---> sin R=7.5*sin 43/5.9
sin R=0.8669---------> R=arc sin (0.8669)----> R=60.11°----> R=60°

the answer Part 1) is the option
D. 60

Part 2) In triangle ABC mA=43, mB=62 and BC=22 in. What is AB to the nearest tenth of an inch?

we know that
∠C=180-(62+43)-----> ∠C=75°

Applying the law of sines
22/sin 43=AB/sin 75------> AB=22*sin 75/ sin 43-----> AB=31.16 in
AB=31.2 in

the answer Part 2) is
D.31.2

You might be interested in

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$

We set f''(x) = 0

$f''(x) =12x+6 =0\\\\x=-\frac{1}{2}$

Analyzing concavity, we get

$\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right$

The function is concave upward on $(-1/2,\infty)$ because the f''(x) > 0 and concave downward on $(-\infty,-1/2)$ because the f''(x) < 0.

f(x) is concave down before $x = -\frac{1}{2}$ , concave up after it. So f(x) has an inflection point at $x = -\frac{1}{2}$ .

7 0

3 years ago

Write an expression y+3y-2y=

nasty-shy [4]

Answer:

3y=2y

Step-by-step explanation:

Im not sure if this is the answer your looking for but I hope this helped. I tried tho lol :)

4 0

3 years ago

Read 2 more answers

Assume that blood pressure readings are normally distributed with a mean of 123 and a standard deviation of 9.6. If 144 people a

Vanyuwa [196]

Answer:

The probability is 0.9938

Step-by-step explanation:

In this question, we are asked to calculate the probability that the mean blood pressure readings of a group of people is less than a certain reading.

To calculate this, we use the z score.

Mathematically;

z = (mean - value)/(standard deviation/√N)

From the question, we can identify that the mean is 125, the value is 123 , the standard deviation is 9.6 and N ( total population is 144)

Let’s plug these values;

z = (125-123)/(9.6/√144) = 2.5

Now we proceed to calculate the probability with a s score less than 2.5 using statistical tables

P(z<2.5) = 0.9938

8 0

3 years ago

12-1(23.28)34.67+123=? çözümü

xxMikexx [17]