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3 years ago

Find the length of the missing side of the right triangle below.

Mathematics

1 answer:

Elena-2011 [213]3 years ago

8 0

Answer:

your answer is a = <u>+</u> √105 .

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Y-4x=3. 2x-3y=21 can someone help

mars1129 [50]

Y-4X =3 ; 2x-3y=21

Y= 3+ 4x

2x -3( 3+4x)= 21

2x- 9- 12x= 21

-10x -9= 21

-10x -9+9= 21+9

-10x = 30

X= 30/-10= -3

Y= 3+4x = 3+ 4* (-3)= 3-12= -9

3 0

3 years ago

What’s the inequality?

Hitman42 [59]

Answer:

x: numbers of hot chocolate selling

y: numbers of apple cide selling

1x + 2y ≥ 240

2y ≥ 240 - x

y ≥ 120 - 0.5*x

7 0

2 years ago

Say there are 2 candles per every 20 cakes. How many candles would there be for every 60 cakes?

abruzzese [7]

Answer:

There would be 6 candles.

Step-by-step explanation:

So you said that there would be 2 candles per 20 cakes, so I simplified it to 1 candle per 10 cakes. So, if there were 60 cakes, there would be 6 candles.

I hope it helped!

~Flipohrama

6 0

2 years ago

Read 2 more answers

(11/x^2-25)+(4/x+5)= 3/x-5

Ierofanga [76]

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\
\cfrac{11}{x^2-25}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\implies \cfrac{11}{x^2-5^2}+\cfrac{4}{x+5}=\cfrac{3}{x-5}$

$\bf \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\impliedby 
\begin{array}{llll}
\textit{notice, LCD is }(x-5)(x+5)\\
\textit{so let's multiply all by it}\\
\textit{to toss the denominators}
\end{array}$

$\bf (x-5)(x+5)\left( \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5} \right)=(x-5)(x+5)\left( \cfrac{3}{x-5} \right)
\\\\\\
11+4(x-5)=3(x+5)\implies 11+4x-20=3x+15
\\\\\\
4x-9=3x+15\implies x=24$

8 0

3 years ago

A company's paper shredder holds 35 ft 3 of shredded paper the company must empty the shredded paper into 8 equal-sized bags for

Vitek1552 [10]

Multiply 35 by its self three times and then divide it by eight

8 0

3 years ago