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IrinaK [193]
3 years ago
10

What is the answer? Brainliest and 13 pointsssssss

Mathematics
1 answer:
vodomira [7]3 years ago
5 0

Answer:

33

Step-by-step explanation:

2x^2 +3y

Let x =3 and y=5

2 *(3)^2 +3(5)

We do exponents first

2 * 9  + 3(5)

Then multiply

18+ 15

Add

33

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A ladder is placed 7 feet from the base of a building and reaches a point on the building that is 24 feet above the ground.Find
Alinara [238K]
◆ Pythagoras Theoram ◆

Hey !!

Check the attachment.
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5 0
2 years ago
Find the volume of the solid formed by revolving the region bounded by LaTeX: y = \sqrt{x} y = x and the lines LaTeX: y = 1 y =
Strike441 [17]

Answer:

The volume is:

\displaystyle\frac{37\pi}{10}

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

\displaystyle\int_a^b\pi r^2dx

Notice here the radius of the washer is the difference of the given curves:

x-\sqrt{x}

So the integral becomes:

\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx

We solve it:

Factor \pi out and distribute the exponent (you can use FOIL):

\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx

Notice: x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}

So the integral becomes:

\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx

Then using the basic rule to evaluate the integral:

\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4

Simplifying a bit:

\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4

Then plugging the limits of the integral:

\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]

Taking the root (rational exponents):

\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]

Then doing those arithmetic computations we get:

\displaystyle\frac{37\pi}{10}

6 0
3 years ago
Area of wrapping paper 5.5 in by 2.1 in
ExtremeBDS [4]
Multiply 5.5 times 2.1 
answer is 11.55in
6 0
3 years ago
Read 2 more answers
= 1/2 (u+v)t
RoseWind [281]

The distance travelled by an object is 50 m.

<h3>What is Distance?</h3>

The distance between two points is the length of the path connecting them.

Here, given equation of distance

         S = 1/2 (u + v)t

Where s is the distance traveled by an object (in meters)

initial velocity u (in m/s)

final velocity v (in m/s).

t in seconds.

Now, given values are;

u = 8 m/s ;  v = 12 m/s ;   t = 5 sec.

    S = 1/2 (8 + 12)5

    S = 1/2 (20)5

    S = 100/2

    S = 50 m.

Thus, the distance travelled by an object is 50 m.

Learn more about Distance from:

brainly.com/question/15172156

#SPJ1

5 0
2 years ago
Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a
8_murik_8 [283]

Answer:

  • a) 3/5·((-2)^n + 4·3^n)
  • b) 3·2^n - 5^n
  • c) 3·2^n + 4^n
  • d) 4 - 3 n
  • e) 2 + 3·(-1)^n
  • f) (-3)^n·(3 - 2n)
  • g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form a[n]=r^n, then it will satisfy ...

  r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}

Rearranging and dividing by r^{n-2}, we get the quadratic ...

  r^2-c_1r-c_2=0

The quadratic formula tells us values of r that satisfy this are ...

  r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

  a[n]=pr_1^n+qr_2^n

We can find p and q by solving the initial condition equations:

\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

These have the solution ...

p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}

_____

Using these formulas on the first recurrence relation, we get ...

a)

c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

  a[n]=(p+qn)r^n

The initial condition equations are now ...

\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

and the solutions for p and q are ...

p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}

__

Using these formulas on problem (d), we get ...

d)

c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n

__

And for problem (f), we get ...

f)

c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0
2 years ago
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