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3 years ago

1.Graph the functions on the same coordinate plane.

1 answer:

4 0

1. solutions are intersection points
they intersect at x=4 and x=1 or the points (4,0) and (1,0)
select 4 and 1

2.
sub -x for x
if you get the same function, it is even
if you get the exact negative of the function, it is odd
if neither, then neither
basically
f(-x)=f(x) is even
f(-x)=-f(x) is odd
and neither is neighter

so
f(-x)=-2(-x)²+3(-x)
f(-x)=-2(x)²-3x
f(-x)=-2x²-3x
that is not the same function nor is it the exact oposite
neither even nor odd

3.
domain is the time you can use
obvioulsy start at time=0
and stop when it hits the ground because it shouldn't go underground
so domain is all real numbers from 0 to 5 including 0 and 5
that is [0,5]

4. (the (f+g)(x) one)
(f+g)(x)=f(x)+g(x)
so that is just
x²-36+x³+2x²-10=
x³+3x²-46

5. just mulitply them
x⁷+9x⁴-9x³-81

6. minus them
(3x⁵+6x²-5)-(2x⁴+7x²-x+16)
3x⁵+6x²-5-2x⁴-7x²+x-16
3x⁵-2x⁴-x²+x-21

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1)The square shown has a perimeter of 32 units. The square is rotated about line k.

natali 33 [55]

1. The perimeter of the square is 32 units, means each side is 32/4=8 units.

rotating the square about line k forms a cylinder with base radius equal to 8 units, and height equal to 8 units. (check the first picture)

the circumference of the base is C=2πR=16π units=16*3.14 units = 50.24 units.

Answer: a cylinder with a circumference of about 50 units

2.
Rotating the equilateral triangle around line a, produces a cone with base radius equal to half of one side of the triangle, that is r=10mm.

(check picture 2)

The circumference is C = 2πr=C = 2*3.14*10=62.8 mm

Answer: a cone with a base circumference of about 63 mm

5 0

3 years ago

Read 2 more answers

A collection of coins consists of nickels, dimes, and quarters. There are three fewer quarters than nickels and six more dimes t

weeeeeb [17]

Answer:

You have 15 dimes. You have 9 quarters. You have 12 nickels

Step-by-step explanation:

lets set some variables:

let "n" = the number of nickels

let "d" = the number of dimes

let "q" = the number of quarters

So, the total amount of money you have should be: $4.35 = 0.25q + 0.10d + 0.05n

Now let's look at the relationships between the coins:

"There are three fewer quarters than nickels": n - 3 = q

"six more dimes than quarters": q + 6 = d

So now you have three equations with three variables, all you need to do is solve.

$\left \{ {{4.35 = 0.25q + 0.10d + 0.05n} \atop {n - 3 = q}} \atop {q + 6 = d}}\right.$

first, you can substitute "n-3" for "q" (according to the 2nd equation) in the 1st and 3rd equation, you get:

$\left \{ {{4.35=0.25(n-3)+0.10d+0.05n} \atop {(n-3)+6=d}} \right.$

You now only have two equations and two variables.

Simplify:

$\left \{ {{4.35=0.25n-0.75+0.10d+0.05n} \atop {n+3=d}} \right.$

$\left \{ {{4.35=0.30n-0.75+0.10d} \atop {n+3=d}} \right.$

Now substitute "n+3" for "d" (according to the 2nd equation) in the 1st equation:

4.35=0.30n-0.75+0.10(n+3)

simplify:

4.35=0.30n-0.75+0.10n+0.30

4.35=0.40n-0.45

4.35+0.45=0.40n

4.80=0.40n

n=12

You have 12 nickels. Now sub "n" back into your equations to find the number of dimes and quarters:

n - 3 = q

12 - 3 = q

q = 9

You have 9 quarters.

q + 6 = d

9 + 6 = d

d = 15

You have 15 dimes.

3 0

3 years ago

For which values of the parameter B the following system admits a unique solution [ 1 b 1-b 2 2 0 2–2B 4 0] [x y z] = [1 2 0]

photoshop1234 [79]

Answer: β ≠ ±1

Step-by-step explanation: For a system of equations to have an unique solution, its determinant must be different from 0: det |A| ≠ 0. So,

det $\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right]$ ≠ 0

Determinant of a 3x3 matrix is calculated by:

det $\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right]\left[\begin{array}{ccc}1&\beta\\2&2\\2-2\beta&4\end{array}\right]$

$8(1-\beta)-[2(2-2\beta)(1-\beta)]$

$8-8\beta-4+8\beta-4\beta^{2}$

$-4\beta^{2}+4\neq 0$

$\beta^{2}\neq 1$

$\beta \neq \sqrt{1}$

β ≠ ±1

For the system to have only one solution, β ≠ 1 or β ≠ -1.

7 0

3 years ago

The legth of a rectangle is the same as the legth of each side of a square. The legth rectangle is 4 cm more than 3 times the wi

ankoles [38]

Let, length and breadth of rectangle is L and B respectively.

It is given that :

The length rectangle is 4 cm more than 3 times the width of the rectangle.

L = 3B + 4 ......1 )

Also, area of square = area of the rectangle + 66

L² = LB + 66

Putting value of L from

L² = ( 3B + 4 )( B ) + 66

L² = 3B² + 4B + 66

( 3B + 4 )² = 3B² + 4B + 66

9B² + 16 + 24B = 3B² + 4B + 66

6B² + 20B - 50 = 0

3B² + 10B - 25 = 0

3B² + 15B - 5B -25 = 0

3B( B + 5 ) -5( B + 5 ) = 0

B = 5/3 units

L = 3( 5/3 ) + 4

L = 5 + 4 = 9 units

Hence, this is the required solution.

6 0

3 years ago

Can someone please help me on this question?

yan [13]

Answer: q³⁰

Explanation:

First just solve the first part using the exponent rules

p²q⁵ becomes 1/p-⁸q-²⁰ then we flip the fraction so the exponents become positive. Now we have p⁸q²⁰.

Before multiplying the other equation, we must simplify. p-⁴q⁵ becomes 1/p⁴q-⁵ and since it's the exponents being raised to a power we simply multiply the inner exponents times the outer exponent which yields 1/p⁸q-¹⁰. We must make q-¹⁰ positive so we will then bring it to the numerator of the fraction which gives us: q¹⁰/p⁸.

Multiply q¹⁰/p⁸ * p⁸q²⁰/1 = p⁸q³⁰/p⁸ divide the p exponents by each other which yields 0 since when u divide exponents you just subtract them so 8 - 8 = 0. Your answer is now q³⁰/1 or just q³⁰

4 0

3 years ago