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3 years ago

Please please help if u don’t know please don’t answer not trying to be mean I know u are trying

Mathematics

1 answer:

Gemiola [76]3 years ago

3 0

x= 3

awutydyt7asd67awt67

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CAN SOMEONE PLEASE HELP ME

devlian [24]

Answer:

Step-by-step explanation:

The first two are parallel but the other are not

6 0

3 years ago

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Sofia hits a softball straight up. the equation h= -16t^2 + 90t models the height h, in feet, of the ball after t seconds. How l

andrezito [222]

It is in the air 5.625 seconds.

Using the quadratic formula,

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\\
\\=\frac{-90\pm \sqrt{90^2-4(-16)(0)}}{2(-16)}
\\
\\=\frac{-90\pm \sqrt{8100-0}}{-32}
\\
\\=\frac{-90\pm \sqrt{8100}}{-32}=\frac{-90\pm90}{-32}
\\
\\=\frac{-90+90}{-32}\text{ or }\frac{-90-90}{-32}=\frac{0}{32}\text{ or }\frac{-180}{-32}
\\
\\=0\text{ or }5.625$

0 is when the ball is first launched, 5.625 is when it hits the ground again.

4 0

3 years ago

Find the value for a (geometry)

coldgirl [10]

15 degrees

I hope it will be answer of your question

4 0

3 years ago

Find the angle between u = (8.- 3) and v = (-3,- 8) Round to the nearest tenth of a degree.

Nimfa-mama [501]

Answer:

Step-by-step explanation:

First you must calculate the module or the magnitude of both vectors

The module of u is:

$|u|=\sqrt{(8)^2 + (-3)^2} \\\\|u|=\sqrt{64 + 9}\\\\|u|=8.544$

The module of v is:

$|v|=\sqrt{(-3)^2 + (-8)^2} \\\\|u|=\sqrt{9 + 64}\\\\|u|=8.544$

Now we calculate the scalar product between both vectors

$u*v = 8*(-3) + (-3)*(-8)\\\\u*v = -24+ 24=0$

Finally we know that the scalar product of two vectors is equal to:

$u*v = |u||v|*cos(\theta)$

Where $\theta$ is the angle between the vectors u and v. Now we solve the equation for $\theta$

$0 = 8.544*8.544*cos(\theta)\\\\0 = cos(\theta)\\\\\theta= arcos(0)\\\\\theta=90\°$

the answer is 90°

Whenever the scalar product of two vectors is equals to zero it means that the angle between them is 90 °

5 0

4 years ago

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write the equation of a line in slope intercept form that contains the points ( - 5,2 ) and ( 1, - 4 )

Natalija [7]

Equation of a line is given by y - y1 = m(x - x1); where m = (y2 - y1)/(x2 - x1)

y - 2 = ((-4 - 2)/(1 - (-5))(x - (-5))
y - 2 = (-6/(1 + 5))(x + 5)
y - 2 = (-6/6)(x + 5)
y - 2 = -1(x + 5)
y - 2 = -x - 5
y = -x - 5 + 2
y = -x - 3

8 0

3 years ago