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3 years ago

I’m confused please help

Mathematics

1 answer:

Law Incorporation [45]3 years ago

3 0

So x equals 30 because u did 180-150 and y is also 30 since you found x you then add 90 plus 30 and it will give u 120 then 180 minus 120 will be 60 then 90 - 60 because u are trying to find y, and u will get 30.

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Please anyone can help me to solve the questions please

Dafna11 [192]

Answer:

144 169 225

Step-by-step explanation:

12² = 144

13² = 169

15² = 225

5 0

3 years ago

Read 2 more answers

Yellow Cab Taxi charges a $1.75 flat rate in addition to $0.65 per mile. Katie has no more than $10 to spend on a ride. How many

spayn [35]

About miles.

Subtract the flat rate of 1.75 from the spending limit of 10$ which leaves you with 8.25 then divide the rest of your spending money (8.25$) into your rate per mile (0.65) , so essentially , 8.25 / 0.65 which leaves you with the answer of ;
12 miles.

3 0

2 years ago

Read 2 more answers

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

4. Make up a system of equations with infinite<br> solutions

Taya2010 [7]

Answer:

$\left \{ {{x+y=1} \atop {2x+2y=2}} \right.$

Step-by-step explanation:

As long as the two equations represent the same straight line on the coordinate plane (they overlap), there will be infinite many intersections and infinite number of solutions.

6 0

2 years ago

How can I solve this?!

Nutka1998 [239]

evaluate f(5) and f(2)

f(5) = 5m + b and f(2) = 2m + b, hence

f(5) - f(2) = 5m + b - 2m - b = 3m

the expression simplifies to

$\frac{3m}{3}$ = 2 ( cross- multiply )

3m = 6 ( divide both sides by 3 )

m = 2 → A

8 0

3 years ago