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leonid [27]
3 years ago
9

What is the answer to 3k-1=7k+2

Mathematics
2 answers:
Phoenix [80]3 years ago
5 0
Answer is given above

Paladinen [302]3 years ago
3 0
The answer for 3k-1=7k+2 is -3/4, or -0.75 because:

 3k-1=7k+2
    +1      +1
 3k=7k+3
-7k -7k
-4k=3
/-4  /-4
 k=-3/4, or -0.75
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Carmen paid $17.64 for a 7.94-ky bag of dog food. A few weekel later, she paid $18.88 for an 8,39 ky bag at a different store,
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3 years ago
Find the measurement of the inscribed angle CDB indicated below
Wewaii [24]

Answer:

∠ CDB = 45°

Step-by-step explanation:

The measure of an inscribed angle is half the measure of its intercepted arc.

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7 0
3 years ago
PLEASE HELP! I AM REALLY CONFUSED! find the value of a if X^3+8x^2+ax-2 is divisible by x-1
WARRIOR [948]

The value of a from the equation is -63.

Here we have to find the value of a.

Given data:

The equation is = x³ + 8x² + ax -2

The equation is divisible by x -1

So we have:

x -1 = 0

x = 1

Now putting the value in the equation we get the remainder as o because the equation is divisible by x -1.

So;

1³ + 8² + a(1) - 2 = 0

63 + a = 0

a = -63

Therefore we get the value of a as -63.

To know more about the equation refer to the link given below:

brainly.com/question/14107099

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4 0
1 year ago
Using Cramer’s Rule, what is the value of y in the system of linear equations below?
Vadim26 [7]

for the system of equations

a_1x+b_1y=c_1 \\a_2x+b_2y=c_2

the three matrices needed to use Cramer's Rule are:

D=\left[\begin{array}{cc}a_1&b_1\\a_2&b_2\end{array}\right] \\\\D_x=\left[\begin{array}{cc}c_1&b_1\\c_2&b_2\end{array}\right] \\\\D_y=\left[\begin{array}{cc}a_1&c_1\\a_2&c_2\end{array}\right].

To use Cramer's Rule we have to calculate the three determinants listed below of the matrices listed below.

D=\left[\begin{array}{cc}2&5\\-3&-2\end{array}\right] \\\\D_x=\left[\begin{array}{cc}-13&5\\3&-2\end{array}\right] \\\\D_y=\left[\begin{array}{cc}2&-13\\-3&3\end{array}\right] .

The value of the determinants are shown below.

det(D)=(2)(-2)-(-3)(5)=-4+15=11\\det(D_x)=(-13)(-2)-(3)(5)=26-15=11\\det(D_y)=(2)(3)-(-3)(-13)=6-39=-33\\

The value of y is \frac{det(D_y)}{det(D)}=-\frac{33}{11} =-3..

3 0
4 years ago
Please help me with the answer
gtnhenbr [62]
Answer is D

\frac{4}{x+3}* \frac{x^2+6x+9}{16}=   \frac{1}{(x+3)}* \frac{(x+3)^2}{4}=   \frac{x+3}{4}
7 0
3 years ago
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