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3 years ago

Jennifer 4 years older than her sister, Rebecca. Their ages sum to 28. How old are each of the sisters?

Mathematics

1 answer:

Musya8 [376]3 years ago

8 0

Answer:

Rebecca is 12

Jennifer is 16

Step-by-step explanation:

28 - 4 = 24

24 : 2 = 12

12 + 4 = 16 (Jennifer)

28 - 16 = 12 (Rebecca)

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Y-x^2=5 written in function notation

nataly862011 [7]

Answer:

f(x) = x² + 5

General Formulas and Concepts:

Algebra I

Equality Properties
Function Notation

Step-by-step explanation:

Step 1: Define

y - x² = 5

Step 2: Rewrite

Add x² to both sides: y = x² + 5
Rewrite y: f(x) = x² + 5

7 0

3 years ago

If the discriminant of a quadratic equation is 10, how many solutions does the equation have?

Sunny_sXe [5.5K]

Answer:

C. 2

Step-by-step explanation:

If the discriminant is >0 we have 2 real solutions

If the discriminant is =0 we have 1 real solutions

If the discriminant is <0 we have 2 complex solutions (no real solutions)

since 10>0 we have 2 real solutions

3 0

3 years ago

Find the value of the determinant using the method of expansion by minors; expand on the third row

valkas [14]

For the matrix

$\begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}$

the determinant using the method of expansion by minors, expanding on the third row is:

$\det \begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}=a_{31}\det \begin{bmatrix}{a_{12}} & {a_{13}} & {} \\ {a_{22}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}-a_{32}\det \begin{bmatrix}{a_{11}} & {a_{13}} & {} \\ {a_{21}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}+a_{33}\det \begin{bmatrix}{a_{12}} & {a_{12}} & {} \\ {a_{22}} & {a_{22}} & {} \\ {} & {} & {}\end{bmatrix}$

Answer:

First, we compute the determinants of the minors:

$\begin{gathered} \det \begin{bmatrix}{0} & {4} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=0+4=4, \\ \det \begin{bmatrix}{1} & {4} & {} \\ {1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=3-4=-1, \\ \det \begin{bmatrix}{1} & {0} & {} \\ {1} & {-1} & {} \\ {} & {} & {}\end{bmatrix}=-1-0=-1. \end{gathered}$

Therefore:

$\det \begin{bmatrix}{1} & {0} & {4} \\ {1} & {-1} & {3} \\ {0} & {5} & {-2}\end{bmatrix}=0\times4-5\times(-1)+(-2)\times(-1)=5+2=7.$

3 0

1 year ago

Determine the domain of the function.

AnnyKZ [126]

It it is the first choice i mentioned then B is the answer.

7 0

4 years ago

Read 2 more answers

the base of a ten foot ladder stands six feet from a house. How many feet up the house does the ladder reach

Ludmilka [50]

In this problem you will need to use the Pythagorean theorem (c^2=a^2+b^2).
The a and b represents the two edges, while c is the diagonal side and it is called the hypotenuse. Since you already know what the hypotenuse is and what one of the sides already are you just have to use the problem: c^2-a^2=b^2. Then if you plug the data you already have into the problem you will get 10^2-6^2=b^2. That then equals 100-36=b^2. Then you subtract and get b^2=64. Then you square root both sides and you get the answer b=8.

5 0

4 years ago