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3 years ago

9 gal 1 qt - 5 gal 3 qt

Mathematics

1 answer:

dmitriy555 [2]3 years ago

5 0

If it’s in that order then it would be 4 gal -2 quart, but if it’s not to be in an order it would be 4 gal and 2 qt

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Can someone help me with this problem? Please.

Mamont248 [21]

Answer:

(7x-1)-3(7x-1)

Step-by-step explanation:

First write out the problem

7x^2-22x+3

Next, Factor the expressions

7x^2 - x - 21x + 3

Now factor the whole thing

x(7x-1)-3(7x-1)

4 0

3 years ago

the equation f=v+at represents the final velocity of an object f, with an initial velocity,v, and an acceleration rate a,over ti

vfiekz [6]

F=v+at
f-v=at
(f-v)/t =a
the first choice is correct :)

3 0

3 years ago

Will give Brainlys help with math problem

d1i1m1o1n [39]

Answer:

8 / 15

Step-by-step explanation:

By Pythagoras theorem,

OP² + PQ² = OQ²

OP² + 8² = 17²

OP² = 17² - 8²

= 289- 64

OP² = 225

OP² = 15²

OP = 15

Formula : -

Tan θ = Opposite side / Adjacent side

Here,

θ = Q

Opposite side = PQ = 8

Adjacent side = OP = 15

Tan Q = 8 / 15

4 0

3 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

Simplify the expression 9m-7m+2m

tatuchka [14]

Simplify step-by-step.

9m−7m+2m

=9m+−7m+2m

Combine Like Terms:

=9m+−7m+2m

=(9m+−7m+2m)

=4m

4 0

3 years ago