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inna [77]
3 years ago
11

Fill in the table using this function rule. y=-2x +4

Mathematics
1 answer:
Anton [14]3 years ago
3 0

Answer

i think this is what you mean

Step-by-step explanation:

x          y

2         0

1          2

0         4

-1         6

-2        8

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3x + 5 equals 19 - 4x what does x equals to
GenaCL600 [577]

Answer:

<h3>x = 2</h3>

Step-by-step explanation:

3x + 5 = 19 - 4x     <em>subtract 5 from both sides</em>

3x = 14 - 4x      <em>add 4x to both sides</em>

7x = 14    <em>divide both sides by 7</em>

x = 2

8 0
3 years ago
Read 2 more answers
Mr. Thomas drove 75 miles in May. He drove 6 times as many miles in July as he did in May. He
harkovskaia [24]

Answer:

1800 miles

Step-by-step explanation:

No. of miles driven by Mr. Thomas in May = 75

It is given that miles driven in July is 6 times of miles driven by Mr. Thomas in May(75 miles).

Thus

No. of miles driven by Mr. Thomas in July = 6 * No. of miles driven by Mr. Thomas in May   = 6*75 = 450 miles.

__________________________________________________

Another condition given   that miles driven in June is 4 times of miles driven by Mr. Thomas in July(450miles as calculated above).

Thus

No. of miles driven by Mr. Thomas in June = 4 * No. of miles driven by Mr. Thomas in July   = 4* 450 miles = 1800 miles.

No. of miles driven by Mr. Thomas in June is 1800 miles.

8 0
3 years ago
Write another fraction that is equivalent to 4/5. Draw a diagram to show that they are equal. Then find the equivalent decimals
MaRussiya [10]

8/10 is an equivalent fraction

3 0
3 years ago
How much longer is a 1inch button than a 3/8 inch button
neonofarm [45]
5/8 inches longer since 8/8 - 3/8 = 5/8
4 0
3 years ago
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9966 [12]

Answer:

Step-by-step explanation:

An eigenvalue of n × n is a function of a scalar \lambda  considering that there is a solution (i.e. nontrivial) to an eigenvector x of Ax =  

Suppose the matrix A = \left[\begin{array}{cc}-1&-1\\2&1\\ \end{array}\right]

Thus, the equation of the determinant (A - \lambda1) = 0

This implies that:

\left[\begin{array}{cc}-1-\lambda &-1\\2&1- \lambda\\ \end{array}\right] =0

-(1 - \lambda^2 ) + 2 = 0

-1 + \lambda ^2 + 2= 0

\lambda^2 +1 =0

Hence, the eigenvalues of the equation are \mathtt{\lambda = i , -i}

Also, the eigenvalues can be said to be complex numbers.

3 0
3 years ago
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