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DanielleElmas [232]
2 years ago
14

Find x and y. please help​

Mathematics
1 answer:
anygoal [31]2 years ago
8 0

Answer:

x = 4 \sqrt{2}

y = 4 \sqrt{6}

Step-by-step explanation:

\sin 45 \degree =  \frac{x}{8}  \\  \\ x = 8  \sin 45 \degree \\  \\ x = 8 \times  \frac{1}{ \sqrt{2} }  \\  \\ x =  \frac{8 \sqrt{2} }{2}  \\  \\ x = 4 \sqrt{2}  \\  \\   \\  \tan 30 \degree =  \frac{x}{y}  \\  \\  \frac{1}{ \sqrt{3} }  =  \frac{4 \sqrt{2} }{y}  \\  \\ y = 4 \sqrt{6}

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kakasveta [241]

Answer:

thanks

Step-by-step explanation:

8 0
2 years ago
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<img src="https://tex.z-dn.net/?f=b%5E2-c%5E2-10%28b-c%29" id="TexFormula1" title="b^2-c^2-10(b-c)" alt="b^2-c^2-10(b-c)" align=
tino4ka555 [31]
It its to factor b^2-c^2-10b+10c
5 0
3 years ago
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How to evaluate the integral
Papessa [141]
Remember
for F(x) is the antiderivitive of f(x)
\int\limits^a_b {f(x)} \, dx =F(a)-F(b)
so find the antiderivitive of ((x+1)^2)/x

if we expand we get (x^2+2x+1)/x which simplifies to x+2+(1/x)
the anti-deritivive of  x is (1/2)x^2
the antideritiveve of 2 is 2x
the antideritivieve of 1/x is ln|x|

F(x)=(1/2)x^2+2x+ln|x|+C

\int\limits^2_1 { \frac{(x+1)^2}{x} } \, dx =F(2)-F(1)

F(1)=(5/2)+ln1+C
F(2)=6+ln2+C

F(2)-F(1)=6+ln2+C-(5/2+ln1+C)
F(2)-F(1)=(7/2)+ln2
that is the answer
if you want is simplified or expanded it is about 4.1915
8 0
2 years ago
What is the area of the figure
Tomtit [17]
A²+b²=c²
a=b because it has 2 45° angles and one 90 so it is isosceles right triangle

∴ a² + a² = c²
2a²=24²
2a²=576
a²=288
a=√288
a=16.97056275

Therefore. With A=1/2bh
A=(1/2) * 16.97056275 * 16.97056275
   = (1/2) * 288
   = 144

144 ft²
4 0
3 years ago
Hunter and Brian are creating a poster for a project and they need to combine two different size boards to create their project.
shepuryov [24]

Answer:

<em>The maximum area of the second board must be (5x^2+3x-6) square inches.</em>

Step-by-step explanation:

The total area of two different size boards cannot exceed (7x^2-6x+2) square inches.

The area of one board is  (2x^2-9x+8) square inches.

Suppose, the area of the second board is  y square inches.

That means......

y+(2x^2-9x+8)\leq 7x^2-6x+2\\ \\ y\leq (7x^2-6x+2)-(2x^2-9x+8)\\ \\y \leq 7x^2-6x+2-2x^2+9x-8\\ \\ y\leq 5x^2+3x-6

So, the maximum area of the second board must be (5x^2+3x-6) square inches.

7 0
3 years ago
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