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Liula [17]
3 years ago
14

What are the solutions to the equation (m + 1)2 +1 = 5? (1 point)

Mathematics
1 answer:
ale4655 [162]3 years ago
6 0

Answer:

yes c is correct ... go on keep going

BEST OF LUCK

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If Vector u=(5,-7) and v=(-11,3), 2v-6u=_____ and ||2v-6u||≈_____
Sauron [17]
<h2>Answer:</h2>

2\vec{v}-6\vec{u}=(-52,48) \\ \\ ||2\vec{v}-6\vec{u}||=75.28}

<h2>Step-by-step explanation:</h2>

In this problem we have two vectors:

\vec{u}=(5,-7) \ and \ \vec{v}=(-11,3)

So we need to find two things:

2\vec{v}-6\vec{u}

and:

||2\vec{v}-6\vec{u}||

FIRST:

In this case we have the multiplication of vectors by scalars. A scalar is a simple number, so:

2\vec{v}-6\vec{u} \\ \\ Replace \ \vec{v} \ and \ \vec{u} \ by \ the \ given \ vectors: \\ \\ 2(-11,3)-6(5,-7) \\ \\ Multiply \ each \ component \ by \ the \ corresponding \ scalar:\\ \\ (2\times (-11),2\times 3)+(-6\times 5,-6\times (-7)) \\ \\ (-22,6)+(-30,42) \\ \\ Sum \ of \ vectors: \\ \\ (-22-30,6+42) \\ \\ \therefore \boxed{(-52,48)}

SECOND:

If we name:

\vec{w}=2\vec{v}-6\vec{u}

Then, ||2\vec{v}-6\vec{u}|| is the magnitude of the vector \vec{w}. Therefore:

||\vec{w}||=||2\vec{v}-6\vec{u}|| \\ \\ ||\vec{w}||=||(-52,48)|| \\ \\ ||\vec{w}||=\sqrt{(-58)^2+48^2} \\ \\ ||\vec{w}||=\sqrt{3364+2304} \\ \\ ||\vec{w}||=\sqrt{5668} \\ \\ \boxed{||\vec{w}||=75.28}

5 0
3 years ago
Read 2 more answers
Please give me the answer
Anna11 [10]

Answer:

52

Step-by-step explanation:

sorry if it's wrong

5 0
3 years ago
9) Is 0 a solution to 3x + 8 = 5x + 8? Explain or show your reasoning .​
alex41 [277]

Answer:

x=0, yes.

Step-by-step explanation:

3x+8=5x+8

5x-3x+8=8

2x+8=8

2x=8-8

2x=0

x=0/2

x=0

4 0
3 years ago
Read 2 more answers
Suppose a football is kicked with an initial velocity of 82 ft/sec., at an angle of
satela [25.4K]

Answer:

The position P is:

P = 87\^x + 75\^y ft     <u><em> Remember that the position is a vector. Observe the attached image</em></u>

Step-by-step explanation:

The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity s_0 is:

y(t) = y_0 + s_0t -16t ^ 2

Where y_0 is the initial height = 0 for this case

We know that the initial velocity is:

82 ft/sec at an angle of 58 ° with respect to the ground.

So:

s_0 = 82sin(58\°) ft/sec

s_0 = 69.54 ft/sec

Thus

y(t) = 69.54t -16t ^ 2

The height after 2 sec is:

y(2) = 69.54 (2) -16 (2) ^ 2

y(2) = 75\ ft

Then the equation that describes the horizontal position of the ball is

X(t) = X_0 + s_0t

Where

X_ 0 = 0 for this case

s_0 = 82cos(58\°) ft / sec

s_0 = 43.45 ft/sec

So

X(t) = 43.45t

After 2 seconds the horizontal distance reached by the ball is:

X (2) = 43.45(2)\\\\X (2) = 87\ ft

Finally the vector position P is:

P = 87\^x + 75\^y ft

6 0
3 years ago
Calculate:<br> √5/2-√5+2/2+√5
klio [65]
-√5/2+√5+1
-√5/2+2√5/2+1
√5/2+1
(√5+2)/2

4 0
3 years ago
Read 2 more answers
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