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3 years ago

What is the maximum number of possible solutions for the system shown below?

1 answer:

7 0

$\bf \begin{cases} 3x^2+y^2=4\\[-0.5em] \hrulefill\\ x^2+y=5\implies \boxed{y}=5-x^2 \end{cases} \\\\\\ \stackrel{\textit{substituting \boxed{y} in the 1st equation}}{3x^2+\left( \boxed{5-x^2} \right)^2=4}\implies 3x^2+25-10x^2+x^4=4 \\\\\\ x^4-8x^2+21=0\impliedby \begin{array}{llll} \textit{the degree of this polynomial}\\ \textit{is 4, and by the fundamental}\\ \textit{theorem of algebra, it can}\\ \textit{have \boxed{4} solutions at most} \end{array}$

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Maria, Denise, and Joseph are making cookies for a bake sale. Maria has 4 times as toany okies for the

ElenaW [278]

Answer:

Maria made 248 cookies, Denise made 62 cookies, Joseph made 372 cookies.

Step-by-step explanation:

Number of cookies made by Maria is m,

made by Denise is d,

made by Joseph is j

We can write the following equations from the word:

m = 4d

j = 6d

m + d + j = 682

We can substitute both 1st and 2nd equation in 3rd and solve for d:

m + d + j = 682

4d + d + 6d = 682

11d = 682

d = 62

Thus,

m = 4(62) = 248

and

j = 6(62) = 372

Maria made 248 cookies, Denise made 62 cookies, Joseph made 372 cookies.

4 0

2 years ago

What cross section results when you cut a cylinder perpendicular to its base?

wariber [46]

The cross section that results, is a triangle.

Step-by-step explanation:

Cross section is a shape that forms when a slice or a cut is made through a solid.

A solid can have multiple cross sections depending on where you slice the solid.

Cross section perpendicular to the base of a solid will result in triangles.Cross sections parallel to the base of the solid will result in hexagons and so on.

Cross section help in representing three dimension objects in a two dimension way.

3 0

2 years ago

The Highest Common Factor of three numbers is 6 and the Lowest Common Multiple is 24 What are the three numbers?(PLEASE HELP)

vazorg [7]

Answer:

GCF of 3 and 6 is 3.

LCM of 3 and 6 is 6.

Step-by-step explanation:

6 0

2 years ago

An instructor gives her class the choice to do 7 questions out of the 10 on an exam.

Maksim231197 [3]

Answer:

(a) 120 choices

(b) 110 choices

Step-by-step explanation:

The number of ways in which we can select k element from a group n elements is given by:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:

$10C7=\frac{10!}{7!(10-7)!}=120$

Then each student have 120 possible choices.

On the other hand, if a student must answer at least 3 of the first 5 questions, we have the following cases:

1. A student select 3 questions from the first 5 questions and 4 questions from the last 5 questions. It means that the number of choices is given by:

$(5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50$

2. A student select 4 questions from the first 5 questions and 3 questions from the last 5 questions. It means that the number of choices is given by:

$(5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50$

3. A student select 5 questions from the first 5 questions and 2 questions from the last 5 questions. It means that the number of choices is given by:

$(5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10$