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3 years ago

Help me pls? :) ok thx

Mathematics

1 answer:

marishachu [46]3 years ago

8 0

1 hour equals 70 miles so 70 times 4 equals 280 and half of 70 is 35 since you traveled 4.5 hours and 280+35 equals 315

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What is the solution to the system of equations?<br><br> y = 2x + 4<br> y = -x + 1

den301095 [7]

Step-by-step explanation:

start by setting each equation to each other

$2x + 4 = - x + 1$

first we are going to subtract 2x from both side of the equation

$4 = - 3x + 1$

now subtract 1 from both sides

$3 = - 3x$

now divide by -3 on both sides

5 0

3 years ago

10 · c − 8 · 95 = (10 − 8) · 95

Westkost [7]

Answer:

x=95

Step-by-step explanation:

First, we can simplify the equation. 10c-760=190. We can +760 on both sides. 10c=950. We can do 950/10=95.

6 0

2 years ago

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

The council rates are set to rise by 15%. If Teresa pays $950 per year now, what will she pay next year after the rise?

Eddi Din [679]

Answer:

She will pay $1092.50 because .15x950 is 142.50 and if you add that to 950. you get $1092.50

7 0

3 years ago

I just need help with the equation part

guajiro [1.7K]

Answer:

just list the number of albums times the amount they cost plus songs and how much they cost and you got your equation

Step-by-step explanation:

3 0

3 years ago