All categories

3 years ago

Find the approximate value of b in the triangle shown.131285

Mathematics

2 answers:

Paul [167]3 years ago

8 0

Answer:

The side b is 5 units

Step-by-step explanation:

Given the two sides of triangle and one angle of triangle we have to find the remaining side of triangle.

a=12 units

c=10 units

∠B=24°

we have to find the side b of triangle ABC

By cosine law,

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos 24^{\circ}=\frac{12^2+10^2-b^2}{2(12)(10)}=\frac{144+100-b^2}{240}$

$b^2=244-240\cos 24^{\circ}=24.7490901658$

$b=4.97\sim 5units$

givi [52]3 years ago

4 0

Answer:

The answer is D. 5

Step-by-step explanation:

You might be interested in

If g=27 and F=54º, find h. Round to the nearest tenth.

antoniya [11.8K]

Answer:

The correct answer is B = 15.9

Step-by-step explanation:

h = 27 cos 54 = 15.9

4 0

3 years ago

Read 2 more answers

Chris worked 28 hours last month and earned $9 per hour. He spent $18 of his earnings since then.How much money does he have lef

uranmaximum [27]

The answer would be 234. You will multiply 28 by 9 and minus 18.

6 0

3 years ago

Read 2 more answers

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Brilliant_brown [7]

Answer:

$f(x)=\sqrt{x-1}$ -----> Graph C

$f(x)=\sqrt{x}$ -----> Graph A

$f(x)=\sqrt{x}-1$ -----> Graph B

$f(x)=-\sqrt{x}$ -----> Graph D

$f(x)=-\sqrt{x-1}$ -----> Graph E

Step-by-step explanation:

we have

case a) $f(x)=\sqrt{x-1}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x-1\geq 0$

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

The range is the interval -----> [0,∞)

All real numbers greater than or equal to 0

case b) $f(x)=\sqrt{x}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> [0,∞)

All real numbers greater than or equal to 0

case c) $f(x)=\sqrt{x}-1$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> [-1,∞)

All real numbers greater than or equal to -1

case d) $f(x)=-\sqrt{x}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> (-∞,0]

All real numbers less than or equal to 0

case e) $f(x)=-\sqrt{x-1}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x-1\geq 0$

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

The range is the interval -----> (-∞,0]

All real numbers less than or equal to 0

5 0

3 years ago

Solve using elimination.<br> -3x + 4y = 9<br> 3x – 10y = 9<br> -<br> PLEASE HELP

Art [367]

Answer:

x= -7

Y= -3

Step-by-step explanation:

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B10%28x%20-%20y%29%20-%204%281%20-%20x%29%7D%7B3%7D%20%20%3D%20y" id="TexFormula1"

ikadub [295]

so first off, let's simplify both equations, starting off by multiplying both sides by the LCD of all fractions, to do away with the denominators.

$\bf \cfrac{10(x-y)-4(1-x)}{3}=y\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{10(x-y)-4(1-x)=3y} \\\\\\ 10x-10y-4+4x=3y\implies \boxed{14x-13y=4} \\\\[-0.35em] ~\dotfill\\\\ 7+x-\cfrac{x-3y}{4}=2x-\cfrac{y+5}{3}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( 7+x-\cfrac{x-3y}{4} \right)=12\left( 2x-\cfrac{y+5}{3} \right)} \\\\\\ 84+12x-3(x-3y)=24x-4(y+5) \\\\\\ 84+12x-3x+9y=24x-4y-20\implies \boxed{-15x+13y=-124}$

now, let's do some elimination on those two simplified equations.

$\bf \begin{array}{cllcl} 14x&-13y&=&4\\ -15x&+13y&=&-124\\\cline{1-4} -x&&=&-120 \end{array}~\hfill x=\cfrac{-120}{-1}\implies \blacktriangleright x=120 \blacktriangleleft \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{14(120)-13y=4}\implies 1680-13y=4\implies 1680-4=13y \\\\\\ 1676=13y\implies \blacktriangleright \cfrac{1676}{13}=y \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left( 120~,~\frac{1676}{13} \right)~\hfill$

6 0

3 years ago

Find the approximate value of b in the triangle shown.131285​

Find the approximate value of b in the triangle shown.131285