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3 years ago

In the right triangle above, a = 99 and c= 124. Find the value of b, rounded to the nearest tenth.

Mathematics

1 answer:

Murljashka [212]3 years ago

8 0

Answer:

B= -40

Step-by-step explanation:

So a right triangle has to = 180

So we write the given, which 124+99

we are trying to get 180 so lets subtract that by the equation, 180-124+99

180-(124+99)

12+99=223

180-223

=-43

But in your case it has to be rounded to the nearest tenth

So the answer is -40

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Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

2 years ago

The difference between a number and 8 is less than -15

Olenka [21]

Answer:

x-8<-15

Step-by-step explanation:

8 0

3 years ago

Pettit printing company has a total market value of $100 million, consisting of 1 million shares selling for $50 per share and $

amm1812

The weighted average cost of capital for the firm will be 11.25%.

<h3>How to calculate the WACC?</h3>

The weighted average cost of capital is the calculation of the cost of capital for a firm where each category of capital is weighted.

Here, the weighted average cost of capital will be:

= 0.5(10%)(1 - 15%) + 0.5(14%)

= 0.5(0.1)(0.85) + 0.5(0.14)

= 11.25%

The corporate value at 70% debt when WACC is 11.94% will be:

= (EBIT)(1 - T)/WACC

= (13.24)(1 - 0.15)/0.1194

= $94.26 million

The corporate value at 30% debt when WACC is 11.14% will be:

= (EBIT)(1 - T)/WACC

= (13.24)(1 - 0.15)/0.1114

= $101.02 million

Learn more about WACC on:

brainly.com/question/25566972

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4 0

2 years ago

Mr.Robinson ordered 4 cheeseburgers for $2.95 each and 4 sodas for $1.59 each. If he paid with a $20 bill, how much change did h

Fynjy0 [20]

4 x 2.95 = 11.80$
4 x 1.59 = 6.36$
11.80 + 6.36 = 18.16 $
20 - 18.16 = 1.84

Mr. Robinson will get $1.84 in change

5 0

2 years ago

Read 2 more answers

The notation Ak meas the matrix A Multiplied with itself k times (a) For the 2×2 identity matrix I, show that I2 =I (b)For the n

djverab [1.8K]

Answer:

Entries of I^k are are also identity elements.

Step-by-step explanation:

a) For the 2×2 identity matrix I, show that I² =I

$I^{2}=\left[\begin{array}{cc}1&0\\0&1\end{array}\right] \times \left[\begin{array}{cc}1&0\\0&1\end{array}\right] \\\\=\left[\begin{array}{cc}1\times 1+0\times 0&1\times 0+0\times 1\\0\times 1+1\times 0&0\times 0+1\times1\end{array}\right] \\\\=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

Hence proved I² =I

b) For the n×n identity matrix I, show that I² =I

n×n identity matrix is as shown in figure

Elements of identity matrix are

$\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\$

As square of 1 is equal to 1 so for n×n identity matrix I, I² =I

As mentioned above

$\delta I_{ij}=1\quad if\quad i=j\\\delta I_{ij}=0\quad if\quad i\ne j\\$

Any power of 1 is equal to 1 so kth power of 1 is also 1. According to this Ik=I

6 0

3 years ago