<span>The problem is to calculate the angles of the triangle. However, it is not clear which angle you have to calculate, so we are going to calculate all of them
</span>
we know that
Applying the law of cosines
c²=a²+b²-2*a*b*cos C------> cos C=[a²+b²-c²]/[2*a*b]
a=12.5
b=15
c=11
so
cos C=[a²+b²-c²]/[2*a*b]---> cos C=[12.5²+15²-11²]/[2*12.5*15]
cos C=0.694------------> C=arc cos (0.694)-----> C=46.05°-----> C=46.1°

applying the law of sines calculate angle B
15 sin B=11/sin 46.1-----> 15*sin 46.1=11*sin B----> sin B=15*sin 46.1/11
sin B=15*sin 46.1/11-----> sin B=0.9826----> B=arc sin (0.9826)
B=79.3°

calculate angle A
A+B+C=180------> A=180-B-C-----> A=180-79.3-46.1----> A=54.6°

the angles of the triangle are
A=54.6°
B=79.3°
C=46.1°

7 0

2 years ago

A7-ft ladder is placed against a wall with its base 3 ft from the wall. How high above the ground is the top of the ladder? Roun

Firdavs [7]

Answer:

6.3 feet

Step-by-step explanation:

use pythagorean theorem

a² + b² = c²

a² + 3² = 7²

a² + 9 = 49

Subtract 9 from both side

a² = 40

Take the square root of both sides

a = 6.324555320336759

Rounded

a = 6.3

4 0

2 years ago

What is f(g(x)) for x > 5?

mel-nik [20]

Answer:

$\large\boxed{B.\ 4x^2-41x+105}$

Step-by-step explanation:

$f(x)=4x-\sqrt{x}\\\\g(x)=(x-5)^2\\\\f(g(x))\to\text{put}\ x=(x-5)^2\ \text{to}\ f(x):\\\\f(g(x))=f\bigg((x-5)^2\bigg)=4(x-5)^2-\sqrt{(x-5)^2}\\\\\text{use}\\(a-b)^2=a^2-2ab+b^2\\\sqrt{x^2}=|x|\\\\f(g(x)=4(x^2-2(x)(5)+5^2)-|x-5|\\\\x>5,\ \text{therefore}\ x-5>0\to|x-5|=x-5\\\\f(g(x))=4(x^2-10x+25)-(x-5)\\\\\text{use the distributive property:}\ a(b+c)=ab+ac\\\\f(g(x))=(4)(x^2)+(4)(-10x)+(4)(25)-x-(-5)\\\\f(g(x))=4x^2-40x+100-x+5\\\\\text{combine like terms}\\\\f(g(x))=4x^2+(-40x-x)+(100+5)\\\\f(g(x))=4x^2-41x+105$

6 0

3 years ago

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