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Gnom [1K]
4 years ago
10

Can someone help me please

Mathematics
1 answer:
kompoz [17]4 years ago
6 0

Answer: 10 + 6

Step-by-step explanation: 2 negatives cancel to make a positive

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The height of a volleyball, h, in feet, is given by h = −16t^2 + 11t + 5.5, where t is the number of seconds after it has been h
faust18 [17]

Answer:

The volleyball travel high enough to clear the top of the net.

Step-by-step explanation:

The height of a volleyball, h, in feet, is given by h = −16t² + 11t + 5.5, where t is the number of seconds after it has been hit by a player.

Now, for h = 7.3 feet, we can write

7.3 = - 16t² + 11t + 5.5

⇒ 16t² - 11y + 1.8 = 0

Using the quadratic formula we get,

t = \frac{- (- 11) \pm \sqrt{(-11)^{2} - 4(16)(1.8)}}{2(16)}

⇒ t = \frac{11 \pm 2.4}{32}

⇒ t = 0.27 or t = 0.42

Therefore, for the two real positive values of t the volleyball travel high enough to clear the top of the net. (Answer)

5 0
3 years ago
Help plz ill give extra points
Yuliya22 [10]

Answer:

40

Step-by-step explanation:

were

3 0
3 years ago
Read 2 more answers
Each leg of a 45-45-90 triangle has a length of 6 units. What is the length of its hypotenuse?
andreyandreev [35.5K]
Use the Pythagorean theorem:
(\hbox{one leg})^2+(\hbox{the other leg})^2=(\hbox{hypotenuse})^2 \\
6^2+6^2=x^2 \\
6^2 \times 2=x^2 \\
\sqrt{6^2 \times 2}=x \\ \sqrt{6^2} \times \sqrt{2}=x \\ 6\sqrt{2}=x

 
The length of the hypotenuse is 6√2 units.
5 0
3 years ago
Prove that sin3a-cos3a/sina+cosa=2sin2a-1
Sloan [31]

Answer:

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1

Step-by-step explanation:

we are given

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1

we can simplify left side and make it equal to right side

we can use trig identity

sin(3a)=3sin(a)-4sin^3(a)

cos(3a)=4cos^3(a)-3cos(a)

now, we can plug values

\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}

now, we can simplify

\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}

\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}

\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}

now, we can factor it

\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}

\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}

we can use trig identity

sin^2(a)+cos^2(a)=1

\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}

we can cancel terms

=3-4(1-sin(a)cos(a))

now, we can simplify it further

=3-4+4sin(a)cos(a))

=-1+4sin(a)cos(a))

=4sin(a)cos(a)-1

=2\times 2sin(a)cos(a)-1

now, we can use trig identity

2sin(a)cos(a)=sin(2a)

we can replace it

=2sin(2a)-1

so,

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1


7 0
3 years ago
Read 2 more answers
QUICK I NEED HELP!!
natta225 [31]
7 (6) + 14!!! And this needs to be 20 characters long so I’m adding this
4 0
3 years ago
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