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Paul [167]
3 years ago
7

What is the value of k in the equation 10k-7=7k-15

Mathematics
1 answer:
Julli [10]3 years ago
4 0

Answer:

k=-8/3

Step-by-step explanation:

10k-7=7k-15

10k-7k-7=-15

3k-7=-15

3k=-15+7

3k=-8

k=-8/3

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Let ​f(x)=x2+3x−10​. Enter the x-intercepts of the quadratic function in the boxes. x = and x =
SIZIF [17.4K]

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X= 2.

X= -5.

Step-by-step explanation:

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3 0
2 years ago
1). −3p + 6 p =
Katena32 [7]
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4 0
2 years ago
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Which equation requires the division property of equality to be solved?
Ymorist [56]

Answer:

A. 201.6 = 11.2f

Step-by-step explanation:

we solve for each unknown to verify which equality will require the division property:

#A. 201.6 = 11.2f

201.6=11.2f\\\\f=\frac{201.6}{11.2}=201.6\div 11.2\\\\=18

Hence, division property used.

#B. 35.4=Z ÷ 3.1

35.4=Z\div 3.1\\\\35.4=\frac{Z}{3.1}\\\\Z=35.4\times 3.1\\\\=109.74

Hence, multiplication property used.

#C. t/3.2=15.1

\frac{t}{3.2}=15.1\\\\t=15.1\times 3.2\\\\=48.32

Hence, multiplication property used.

#D. 201.6 = 11.2+c

201.6=11.2+c\\\\201.6-11.2=c\\\\c=190.4

Hence, subtractionproperty used.

5 0
3 years ago
Read 2 more answers
Help me please thanks​
horrorfan [7]

Answer:

325m = c

Step-by-step explanation:

Each meeting room has 325 chairs.

Since we don't know how many meeting rooms there are, the number is replaced with the variable, m.

An equation is a problem with an equals sign so you need to put what it would equal.

Since you don't know how many meeting rooms there are you use the variable, c which represents the total number of chairs.

8 0
2 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
2 years ago
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