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KIM [24]
3 years ago
12

Explain how this is wrong or right please thanks

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
5 0
Im Pretty sure you got it right. because what i did was get random numbers and tried to equal it to the numbers such as 15, 16, and 17. Have a good day and ask questions!!

-Kaylie
Nezavi [6.7K]3 years ago
3 0
You got them right:)
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Can someone explain how to do this?
xenn [34]

Answer:

20      <u>6</u>      12      38

<u>5</u>      <u>20</u>      7       <u>32</u>

25     <u>26</u>     <u>19</u>      70

8 0
3 years ago
Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).
lions [1.4K]

Check the picture below.  so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}

(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill

\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}

6 0
2 years ago
Help me I'm giving 20 points <br><br>a²•b⁶ if a =½ and b=2 ?​
exis [7]

Answer:

try 0.5

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Claire Judice
Julli [10]

Answer:

The values of 'x' are -1.2, 0, 0, -4i or 4i.

Step-by-step explanation:

Given:

The equation to solve is given as:

5x^5+6x^4+80x^3+96x^2=0

Factoring x^2 from all the terms, we get:

x^2(5x^3+6x^2+80x+96)=0

Now, rearranging the terms, we get:

x^2(5x^3+80x+6x^2+96)=0

Now, factoring 5x from the first two terms and 6 from the last two terms, we get:

x^2(5x(x^2+16)+6(x^2+16))=0\\x^2(x^2+16)(5x+6)=0

Now, equating each factor to 0 and solving for 'x', we get:

x^2=0\\x=0\ and\ 0\\\\5x+6=0\\x=\frac{-6}{5}=1.2\\\\x^2+16=0\\x^2=-16\\x=\sqrt{-16}=\pm 4i

There are 3 real values and 2 imaginary values. The value of 'x' as 0 is repeated twice.

Therefore, the values of 'x' are -1.2, 0, 0, -4i or 4i.

4 0
3 years ago
Help please please please
den301095 [7]

there is 120 ways to get the word banana

4 0
3 years ago
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