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3 years ago

15

Meg is a veterinarian. She found that 7,or 25% of the dogs she saw this week were boxers. Steve is also a veterinarian. He found

that 50% of the 10 dogs he saw this week were boxers. Does each person need to find the part, the whole, or the percent?
HELP NOW!!!!!!!!!

1 answer:

Alex787 [66]3 years ago

5 0

Answer:

whole

Step-by-step explanation:

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Ok Skylar what is (this)?

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3 years ago

The field hockey coach is purchasing new uniforms for the team. Company A charges a one-time printing fee of $100 and $12 per un

CaHeK987 [17]

$100 - $61
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8 0

3 years ago

Solve nonhomogenous equation: x"+4x=cost

motikmotik

Answer:

$x(t) = Acos 2t + B sin 2t+\frac{cost}{3}$

Step-by-step explanation:

Given is a non homogeneous second degree equation as

$x"+4x=cost$

Auxialary equation is

$m^2+4 =0\\m = 2i, -2i$

Hence general solution is

x = Acos 2t + B sin 2t

Particular integral is = $\frac{cost}{D^2+4}$

Since t has coefficient 1, we substitute

$D^2 =-1\\PI = \frac{cost}{-1+4} =\frac{cost}{3}$

Hence full solution is

$x(t) = Acos 2t + B sin 2t+\frac{cost}{3}$

6 0

3 years ago

Select all the correct answers.<br> In which pairs of matrices does AB = BA?

horsena [70]

In order to multiply a matrix by another matrix, we multiply the rows in the first matrix by the columns in the other matrix (How this is done is shown below)

To determine the pairs of matrices that AB=BA, we will determine AB and BA for each of the options below.

For the first option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-2\times5)+(1\times3)&(-2\times0)+(1\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-10+3&0+2&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&-7&2&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-2)&(5\times0)+(0\times 1)&(3\times1)+(2\times-2)&(3\times0)+(1\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-4&0+2&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&-1&2&\end{array}\right] \\$

∴ AB≠BA

For the second option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-1\times3)+(2\times6)&(-1\times0)+(2\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-3+12&0+-6&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-1)&(3\times0)+(0\times 2)&(6\times1)+(-3\times-1)&(6\times0)+(-3\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+3&0+-6&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&9&-6&\end{array}\right] \\$

Here AB = BA

For the third option

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]; B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times5)+(0\times3)&(1\times0)+(0\times 2)&(-1\times5)+(2\times3)&(-1\times0)+(2\times2)&\end{array}\right]\\AB= \left[\begin{array}{cc}5+0&0+0&-5+6&0+4&\end{array}\right]\\AB = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(5\times1)+(0\times-1)&(5\times0)+(0\times 2)&(3\times1)+(2\times-1)&(3\times0)+(2\times2)&\end{array}\right]\\BA= \left[\begin{array}{cc}5+0&0+0&3+-2&0+4&\end{array}\right]\\BA = \left[\begin{array}{cc}5&0&1&4&\end{array}\right] \\$

Here also, AB=BA

For the fourth option

$A= \left[\begin{array}{cc}1&0&-2&1&\end{array}\right]; B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$

$AB= \left[\begin{array}{cc}(1\times3)+(0\times6)&(1\times0)+(0\times -3)&(-2\times3)+(1\times6)&(-2\times0)+(1\times-3)&\end{array}\right]\\AB= \left[\begin{array}{cc}3+0&0+0&-6+6&0+-3&\end{array}\right]\\AB = \left[\begin{array}{cc}3&0&0&-3&\end{array}\right] \\$

and

$BA= \left[\begin{array}{cc}(3\times1)+(0\times-2)&(3\times0)+(0\times 1)&(6\times1)+(-3\times-2)&(6\times0)+(-3\times1)&\end{array}\right]\\BA= \left[\begin{array}{cc}3+0&0+0&6+6&0+-3&\end{array}\right]\\BA = \left[\begin{array}{cc}3&0&12&-3&\end{array}\right] \\$

Here, AB≠BA

Hence, it is only in the second and third options that AB = BA

$A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right] B= \left[\begin{array}{cc}3&0&6&-3&\end{array}\right] \\$ and $A= \left[\begin{array}{cc}1&0&-1&2&\end{array}\right]B= \left[\begin{array}{cc}5&0&3&2&\end{array}\right] \\$

Learn more on matrices multiplication here: brainly.com/question/12755004

8 0

3 years ago

Read 2 more answers

Consider the equation:

Sergio039 [100]

Step-by-step explanation:

1.

Subtract the coefficient from both sides, keep 55 on the same side.

${x}^{2} - 6x = 55$

Complete the square by dividing the coefficient by two and squaring it.

${x}^{2} - 6x + 9 = 55 + 9$

Use binomial to factor the left side.

$(x - 3) {}^{2} = 64$

2. Solve for x.

$(x - 3) = 8$

$x = 11$

Remeber the square root of 64 is also -8 so

$x - 3 = - 8$

$x = - 5$

So the solutions are -5 and 11

5 0

3 years ago

Read 2 more answers