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Taya2010 [7]
3 years ago
5

Help with answer please

Mathematics
1 answer:
Hatshy [7]3 years ago
3 0
 the answer that is right is A

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How do you find the legs of a right triangle when only given the hypotenuse?
tangare [24]

Answer:

We know that a^2+b^2=c^2. The 45° angle lets us know that y=x (45+45+90=180), so the problem is y^2+x^2=4sqrt of 2

4 \sqrt{2}  =  \sqrt{32 }  \\    \sqrt{32} ^{2}  = 32

So you get y^2+x^2=32, and from there, since we know x and y are equal, you can just divide 32 by 2 then take the square root of that, so the answer should be <em><u>4</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>both</u></em><em><u> </u></em><em><u>x</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>y</u></em>

Step-by-step explanation:

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To solve a system of linear equations graphically we graph both equations in the same coordinate system. The solution to the system will be in the point where the two lines intersect. The two lines intersect in (-3, -4) which is the solution to this system of equations
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7 0
2 years ago
A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s
Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

h(t)=-16t^2+32t+128

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

-16t^2+32t+128=0

The above is a quadratic equation. It can be solved by the formula as follows :

t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here, a = -16, b = 32 and c = 128

t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s

Neglecting negative value, the projectile will land after 4 seconds.

4 0
3 years ago
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