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3 years ago

Work out the value of angle x. (24) х Isosceles

Mathematics

1 answer:

andreev551 [17]3 years ago

5 0

answer:

x=78

since it's an isosceles triangle base angles are equal sum of all angles in traingle is 180 so

24+x+x=180

2x=156

x=78

hope.. this helped <3!!!

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8. Jenny is getting dressed for school,

UkoKoshka [18]

Answer:

$\huge\boxed{\bf\: Probability = \frac{4}{9}}$

Step-by-step explanation:

According to the given information,

Pairs of black pants = 2
Pair of brown pants = 1
Pairs of blue pants = 2

Then, total pairs of pants = 2 + 1 + 2 = 5

Similarly,

Pink T-shirts = 2
Blue T-shirts = 2

Then, total number of T-shirts = 4

This gives us the total nunber of clothing articles with Jenny, i.e.,

Total pairs of pants + Total number of T-shirts

= 5 + 4

= 9

Now, the probability of pulling out a pair of black pants & a blue T-shirt will be:

Total pairs of black pants + Total nunber of blue T-shirts/Total clothing articles

= 2 + 2 / 9

= 4/9

$\rule{150}{2}$

8 0

3 years ago

A square box lid has an area of 50 square inches. Which is the best estimate of the length of one side?

PtichkaEL [24]

Answer: 7.1 inches

Step-by-step explanation:

The area of a square is simply the square of one of its length. Therefore since we are informed that the area of the square box is given as 50 inches².

Then the best estimate of the length of one side will be the square root of 50 which will be:

= ✓50

= 7.1 inches approximately

5 0

3 years ago

Which of the values shown are potential roots of f(x) = 3x3 â€"" 13x2 â€"" 3x 45? Select all that apply.

Verdich [7]

The potential roots of the function are, $\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$

And the accurate root is 3 it can be determined by using rules of the rational root equation.

<h2>Given that,</h2>

Function; $\rm f(x) = 3x^3 - 13x^2 -3x + 45$

<h3>We have to determine,</h3>

Which of the values shown are potential roots of the given equation?

<h3>According to the question,</h3>

Potential roots of the polynomial are all possible roots of f(x).

$\rm f(x) = 3x^3 - 13x^2 -3x + 45$

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

$\rm p=\dfrac{All\ the \ positive}{Negative\ factors \ of\ 45}$

$\rm q=\dfrac{All\ the \ positive}{Negative\ factors \ of\ 3}$

The factor of the term 45 are,

$\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45$

And The factor of 3 are,

$\pm1, \ \pm3$

All the possible roots are,

$\dfrac{p}{q} = \pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$

Now check for all the rational roots which are possible for the given function,

$\rm f(x) = 3x^3 - 13x^2 -3x + 45\\\\ f(1) = 3(1)^3 - 13(1)^2 -3(1) + 45 = 3-13-3+45 = 32\neq 0\\\\ f(-1) = 3(-1)^3 - 13(-1)^2 -3(-1) + 45 =- 3-13+3+45 = 32\neq 0\\\\ f(3) = 3(3)^3 - 13(3)^2 -3(3) + 45 = 81-117-9+45 =0\\\\ f(-3) = 3(-3)^3 - 13(-3)^2 -3(-3) + 45 = -81+117+9+45 =-144\neq 0$

Therefore, x = 3 is the potential root of the given function.

Hence, The potential roots of the function are, $\pm1, \ \pm3, \ \pm5, \ \pm9,\ \pm15, \ \pm45,\ \pm \dfrac{1}{3},\ \pm \dfrac{5}{3}$ .

For more details about Potential roots refer to the link given below.

brainly.com/question/25873992

8 0

2 years ago

Train A arrives at central station on the hour and every 12 minutes. Train B arrives on the hour and every 15 minutes. When do b

uysha [10]

Train A arrives at central station on the hour and every 12 minutes. Train B arrives on the hour and every 15 minutes. When do both trains arrive at thesame time?

I believe that both trains would arrive at the same time on the hour (D).

Hope this helped :)

8 0

3 years ago

Read 2 more answers

What is 5.316 - 1.942 (show ur work)

Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7$

$\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

$=3.374$

Hence the correct answer is 3.374

7 0

2 years ago