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STALIN [3.7K]
3 years ago
7

Leo practices his violin 12.5 hours each week you are so practices singing for 3.5 hours each week if you buy this is the same a

mount of time each week how many hours does your practice in 10 weeks
Mathematics
1 answer:
Gelneren [198K]3 years ago
7 0

Answer: You would spend 160 hours in total of ten weeks.

Step-by-step explanation: Just add 12.5 + 3.5 which = 16. Then multiply 16 times 10 which is 160, and that is your answer.

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How to find slope when there is the y intercept and x-intercept?
natta225 [31]
First graph your x and y intercept then pick any two points on the line.
\frac{rise}{run} = \frac{y2 - y1}{x2 - x1}
that's the equation used to find your slope

Example:

(4,2) (8,6) let's say those are the points you pick from the line

so then you subtract 6-2=4 (6 is your y2 and 2 is your y1)

then subtracts 8-4=4 (8is your x2 and 4 is your x1)

then your left with
\frac{2}{4}
if you can reduce the reduce in this case you can which leaves you with
\frac{1}{2}
and that's how you find the slope
4 0
3 years ago
F(x) = (x - 2)²(x+1)​
Inessa [10]

(x - 2)²(x+1)​

(x-2)(x-2)

multiply the two brackets together

(x)(x)=x^2

(x)(-2)=-2x

(-2)(x)=-2x

(-2)(-2)=4

x^2-2x-2x+4

x^2-4x+4

(x^2-4x+4)(x+1)

multiply the brackets together

(x^2)(x)=x^3

(x^2)(1)=x^2

(-4x)(x)=-4x^2

(-4x)(1)=-4x

(4)(1)=4

x^3+x^2-4x^2-4x+4x+4

Answer:

x^3-3x^2+4

5 0
4 years ago
The square of 9 less than a number is 3 less than the number. What is the number?
dmitriy555 [2]
I think it’s maybe C
8 0
3 years ago
A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ
jonny [76]

Let \vec r(t),\vec v(t),\vec a(t) denote the rocket's position, velocity, and acceleration vectors at time t.

We're given its initial position

\vec r(0)=\langle0,0,10\rangle\,\mathrm m

and velocity

\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}

Immediately after launch, the rocket is subject to gravity, so its acceleration is

\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}

where g=9.8\frac{\rm m}{\mathrm s^2}.

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}

\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}

and

\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m

\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m

\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}

b. The rocket stays in the air for as long as it takes until z=0, where z is the z-component of the position vector.

10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}

c. The rocket reaches its maximum height when its vertical velocity (the z-component) is 0, at which point we have

-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)

\implies\boxed{z_{\rm max}=125,010\,\rm m}

7 0
3 years ago
I need help for solving 4.8(x+4)=2.16
Bas_tet [7]
I hope this helps you



4,8 (x+4)=2,16



48 (x+4)=21,6


x+4=21,6/48


x+4= 0,45


x=0,45-4


x= -3,55


5 0
3 years ago
Read 2 more answers
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