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kirill [66]
3 years ago
14

A line passes through the points (-15, 4) and (5,8). What is its equation in slope-intercept

Mathematics
1 answer:
vovangra [49]3 years ago
7 0

Answer:

y=1/5x+7

Step-by-step explanation:

just checked it on paper

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Marina86 [1]
The answer is D -60.
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In a triangle ABC, if angle A = 53 degree and Angle C=45 degree then find the value of Angle B is:​
murzikaleks [220]

Answer:

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4 0
2 years ago
Solve x2 – 12x + 59 = 0 for x.
mr_godi [17]
59 is a tough bird to deal with; its only factors are 1 and 59.

Thus, forget about factoring.  Instead, use the quadratic formula, or solve the equation by completing the square.

Please note:  x2 is ambiguous.  Please write x^2 to indicate "the square of 2."

Here you have 1x^2 - 12x + 59 = 0, for which a=1, b=-12 and c=59.

Use the quadratic formula:  x=[-b plus or minus sqrt(b^2-4ac)] / (2a)

to find the two roots.  Notice that the "discriminant" b^2 - 4ac will be negative, meaning that your two roots will be "complex."
3 0
3 years ago
There are 295 students in 6th grade at Central Middle School. Must take science. The maximum number of students in the class is
vesna_86 [32]
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7 0
3 years ago
Determine if the set of polynomials {x^2 –2x+1, 2x² + 3x -4,-x2+x+5) is a linearly independent set in P2. Is it a basis for P? W
WITCHER [35]

Answer with Step-by-step explanation:

We are given that a set of polynomials {x^2-2x+1,2x^2+3x-4,-x^2+x+5}

We have to find that given set is a linearly independent set inP_2

and given set is a basis for P_2 or not.

Matrix of given set of polynomials

A=\left[\begin{array}{ccc}1&-2&1\\2&3&-4\\-1&1&5\end{array}\right]

Linearly Independent set :If any row or any column is not a linear combination of other rows or columns then the set is linearly independent set.

Any row or column  is not a  linear combination of other rows or columns.Therefore, given set is  a linearly independent set .

We know that

P_2=x^2

Element of P_2 is of the form

ax^2+bx+c

Every element of p_2 is a linear combination of given set of polynomials.

Hence, given set is linearly independent in p_2 .

If any set is basis for any vector space then it satisfied the following two conditions

1.Given set is linearly independent.

2.Every element of given vector space spanned by the given set.

Given set of polynomials are linearly independent and spanned every element of P_2.

Therefore, given set is  as basis for p_2 because the set is linearly independent and spanned P_2.

7 0
3 years ago
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