MATHEMATICS HIGH SCHOOL

Marge has saved $350 toward the cost of her vacation. The vacation will cost $1,750. What percent of the cost of the vacation has Marge saved?

Answers

Answer 1
Answer: 350x5=1750
100/5=20
converts to 20%
Answer 2
Answer:

Final answer:

Marge has saved 20% of the total cost of her vacation.

Explanation:

Marge's vacation savings can be calculated as a percentage of the total cost, using the formula: (Amount Saved / Total Cost) * 100. In this case, with a total vacation cost of $1,750 and Marge having saved $350, the calculation is ($350 / $1,750) * 100. When computed, it reveals that Marge has successfully saved 20% of the total vacation expense. This means that she has allocated a fifth of the overall cost, demonstrating a commendable financial preparation for her upcoming getaway. Such conscientious savings practices contribute to a more relaxed and enjoyable vacation experience, as Marge has set aside a significant portion of the required funds.

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A pool has 29,000,000 mL how many liters of water can the pool hold

Answers

1000mL=1L so 29,000,000mL=29,000L
Using a ML to L Calculator the result is 29000000mL= 29000.00L

Does anyone know what the answer is

Answers

Answer:

cheating in entrance

Step-by-step explanation:

circumference: 14π

area: 49π

Answer:

nope

Step-by-step explanation:

Three roots of the polynomial equation x4 - 4x3 - 2x2 + 12x + 9 = 0 are 3, -1 and -1. Explain why the fourth root must be a real number. Find the fourth root.

Answers

Hello,

P(x)=x^4-4x^3-2x^2+12x+9=0
P(3)=3^4-4*3^3-2*3²+12*3+9=81-108-18+36+9=0
P(-1)=(-1)^4 - 4*(-1)^3 -2*(-1)²+12*(-1)+9=1+4-2-12+9=0
P'(x)=4x^3-12x²-4x+12
P'(-1)=4*(-1)^3-12*(-1)²-4*(-1)+12=-4-12+4+12=0
(-1) is a double root
Ok 3,-1,-1 are roots.

If the 4th root is not a real but a complex (a+ib), its conjugate will be also a root , there would be 5 roots and not 4


So, the 4th root is real and equal to 3 ( a double root)

x^4-4x^3-2x²+12x+9=0
==> x^4+x^3-5x^3-5x²+3x²+3x+9x+9=0
==>x^3(x+1)-5x²(x+1)+3x(x+1)+9(x+1)=0
==>(x+1)(x^3-5x²+3x+9)=0
==>(x+1)(x^3+x²-6x²-6x+9x+9)=0
==>(x+1)[x²(x+1)-6x(x+1)+9(x+1)]=0
==>(x+1)²(x²-6x+9)=0
==>(x+1)²(x-3)²=0


P(x)=(x+1)²*(x-3)²

The product of (3+2i)and a complex number is (17+7i). The complex number is?

Answers

Hi there! Let x be the complex number. (3+2i)(x)=(17+7i), (x)=(17+7i)/(3+2i)=(17+7i)/(3+2i)*(3-2i)/(3-2i), (17+7i)*(3-2i)/(9+4), (51-34i+21i+14)/13), (65-13i)/13, 13(5-i)/13=(5-i). Therefore, the complex number is (5-i).

Which long division problem can be used to prove the formula for factoring the difference of two perfect cubes?

Answers

Some of the possible options of the questions are;

A) (a - b) | \overline {a^2 + a \cdot b + b^2}

B) (a + b) | \overline {a^2 - a \cdot b + b^2}

C) (a + b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}

D) (a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}

The difference of two perfect cubes has a binomial factor and a trinomial factor

The option that gives the long division problem that can be used to prove the difference of two perfect cubes is option D

D) \underline {(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}}

Reason:

The formula for factoring the difference of twoperfect cubes is presented as follows;

a³ - b³ = (a - b)·(a² + a·b + b²)

Given that a factor of the difference of two cubes is (a - b), and that we

have; (a³ + 0·a·b² + 0·a²·b - b³) = (a³ - b³), both of which are present in

option D, by long division of option D, we have;

{} \hspace {33} a^2 + a \cdot b + b^2\n(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a^2 \cdot b - b^3}\n{} \hspace {33} \underline{a^3 - a^2 \cdot b }\n{} \hspace {55} a^2 \cdot b + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n {} \hspace {55} \underline{a^2 \cdot b - a \cdot b^2}\n{} \hspace {89} a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n{} \hspace {89} \underline{a \cdot b^2 - b^3}\n{}\hspace {89} 0

By the above long division, we have;

(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3} = a² + a·b + b²

Which gives;

(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3} = (a³ + 0·a·b² + 0·a·b² - b³)/(a - b)

We get;

(a³ + 0·a·b² + 0·a·b² - b³)/(a - b) = a² + a·b + b²

(a - b)·(a² + a·b + b²) = (a³ + 0·a·b² + 0·a·b² - b³) = (a³ - b³)

(a - b)·(a² + a·b + b²) = (a³ - b³)

(a³ - b³) = (a - b)·(a² + a·b + b²)

Therefore;

The long division problem that can be used to prove the formula for

factoring the difference of two perfect cubes is

(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}, which is option D

D) (a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}

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Answer:

The correct options, rearranged, are:

Options:

A)(a^2+ab+b^2)/(a-b)\n\nB)(a^2-ab+b^2)/(a+b)\n\nC)(a^3+0a^2+0ab^2-b^3)/(a+b))\n\n D)(a^3+0a^2+0ab^2-b^3)/(a-b)

And the asnwer is the last option (D).

Explanation:

You need to find which long division can be used to prove the formula for factoring the difference of two perfect cubes.

The difference of two perfect cubes may be represented by:

  • a^3-b^3

And it is, as a very well known special case:

  • a^3-b^3=(a-b)(a^2+ab+b^2)

Then, to prove, it you must divide the left side,    a^3-b^3     , by the first factor of the right side,    a-b

Note that, to preserve the places of each term, you can write:

  • (a^3-b^3)=(a^3+0a^2+0ab^2-b^3)

Then, you have:

  • (a^3+0a^2+0ab^2-b^3)=(a-b)(a^2+ab+b^2)

By the division property of equality, you can divide both sides by the same factor, which in this case will be the binomial, and you get:

  • (a^3+0a^2+0ab^2-b^3)/(a-b)=(a^2+ab+b^2)

That is the last option (D).

Let me get some help real quick big fella​

Answers

Answer:

cos B = 9/41

Step-by-step explanation:

Using trig functions, since we have a right angle,

cos theta = adjacent side/ hypotenuse

Cos B = 18/82

Dividing the right side by 2

cos B = 9/41
The answer is A
You have to use the trigonometric ratio as it is a right angle triangle.
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