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Scilla [17]
3 years ago
10

On a 28-question test, there are 2-point questions, 4-point questions, and 5-point questions. The test is worth a total of 100 p

oints. There are twice as many 2-point questions as 5-point questions on the test. How many 2-point questions are on the test?
3.Write a series of equations that could be used to solve this. define your varbibles


4.Solveby any algebraic method. Work must be shown.
Mathematics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer:

The test has 8 questions of 2 points, as well as 16 of 4 points and 4 of 5 points.

Step-by-step explanation:

Since, on a 28-question test, there are 2-point questions, 4-point questions, and 5-point questions, the test is worth a total of 100 points and there are twice as many 2-point questions as 5- point questions on the test, to determine how many 2-point questions are on the test, the following calculation must be performed:

18 x 2 + 9 x 5 + 1 x 4 = 36 + 45 + 4 = 85

16 x 2 + 8 x 5 + 4 x 4 = 32 + 40 + 16 = 88

14 x 2 + 7 x 5 + 7 x 4 = 28 + 35 + 28 = 91

12 x 2 + 6 x 5 + 10 x 4 = 24 + 30 + 40 = 94

10 x 2 + 5 x 5 + 13 x 4 = 20 + 25 + 52 = 97

8 x 2 + 4 x 5 + 16 x 4 = 16 + 20 + 64 = 100

Thus, the test has 8 questions of 2 points, as well as 16 of 4 points and 4 of 5 points.

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Taft CLB Algebra 2 Unit 1: Systems of Linear Equations and Inequalities 2020-2021 / 3 of 13
Tju [1.3M]

Answer:

B + G \leq 15

6B + 10G \geq 90

Step-by-step explanation:

Given

Represent Babysitting with B

Represent Gas Station with G

Total workhours = At most 15

Earnings = At least $90

First, we need to represent the work hours as inequality

B + G  = At most 15

At most 15 means less than or equal to 15.

So, we have:

B + G \leq 15

Next, we represent the earnings as inequality.

6 hours of babysitting is: 6B

10 hours at gas station is: 10G

So:

6B + 10G is at least 90

At least 90 means greater than or equal to 90

So, we have:

6B + 10G \geq 90

8 0
2 years ago
Please im finna fail my NWEA testtttt
julia-pushkina [17]
I think it’s B. since it’s on the adjacent & hypotenuse side it makes it cosine, and the correct way to set it up is B i believe. hopefully i’m not wrong because it could also be D
7 0
3 years ago
Read 2 more answers
PLEASE HELP ME THANKS IF YOU DO HAVE A NICE DAY
notsponge [240]

Answer:

Step-by-step explanation:

Blank 1: 584,000

Blank 2: 1,369,600

I think you just multiply by the number of years.

6 0
3 years ago
Briefly describe scientific notation. how is it useful for writing large and small​ numbers? how is it useful for making​ approx
Alenkasestr [34]
Scientific notation, also called pattern or notation in exponential form, is a way of writing numbers that accommodate values too large (100000000000) or small (0.00000000001) to be written in a conventional manner.

 The use of this notation is based on powers of 104
 Example:
 Big numbers:
 100000000000 = 1 * 10 ^ {11}

 Small numbers:
 0.00000000001 = 1 * 10 ^{-11}

 However, in areas such as physics and chemistry, scientific notation is important to make approximations such as:
 mass of a proton
 mass of an electron and others.
4 0
3 years ago
13. Determine whether B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} is a basis of R3.
Sholpan [36]

Answer:  Yes, the given set of vectors is a basis of R³.

Step-by-step explanation:  We are given to determine whether the following set of three vectors in R³ is a basis of R³ or not :

B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} .

For a set to be a basis of R³, the following two conditions must be fulfilled :

(i) The set should contain three vectors, equal to the dimension of R³

and

(ii) the three vectors must be linearly independent.

The first condition is already fulfilled since we have three vectors in set B.

Now, to check the independence, we will find the determinant formed by theses three vectors as rows.

If the value of the determinant is non zero, then the vectors are linearly independent.

The value of the determinant can be found as follows :

D\\\\\\=\begin{vmatrix} -1& 1 & -1\\ 1 & 0 & 2\\ 1 & 1 & 0\end{vmatrix}\\\\\\=-1(0\times0-2\times1)+1(2\times1-1\times0)-1(1\times1-0\times1)\\\\=(-1)\times(-2)+1\times2-1\times1\\\\=2+2-1\\\\=3\neq 0.

Therefore, the determinant is not equal to 0 and so the given set of vectors is linearly independent.

Thus, the given set is a basis of R³.

5 0
3 years ago
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