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2 years ago

If 1/5 of the remaining blueberries is used to make muffins how many pounds of blueberries are left in the container

Mathematics

1 answer:

nlexa [21]2 years ago

3 0

Answer:

4/5.

Step-by-step explanation:

would you mind looking at my question :)

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Use Gauss-Jordan elimination to solve the following linear system.

marysya [2.9K]

Just put the coefients in to a matrix

1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14

$\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]$
mulstiply 2nd row by -1 and add to 3rd
$\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]$
divde last row by 2
$\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]$
multiply 2rd row by 6 and add to top one
$\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]$
multiply 1st row by -1 and add to 2nd
$\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]$
divide 2nd row by -3
$\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]$
mulstiply 2nd row by -1 and add to 1st row
$\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]$
divide 1st row by -3
$\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]$

rerange
$\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]$

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer

7 0

2 years ago

Write the sentence as an equation. The product of r and 338 plus 97 is the same as r.

Vitek1552 [10]

Answer:

Step-by-step explanation:

To write this sentence as equation

We get

r×338+97=r

338r+97=r

To solve further

r-338r=97

-337r=97

r=-0.2878

8 0

2 years ago

Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos

Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

Length between $P_1$ and $P_{2}$ , (Side 1) :

$(x_1, y_1) = (- 1, 4)$ and

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2$

$= \sqrt{7^2 + 2^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}}$

Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

$(x_1, y_1) = (-1, 4)$

$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

$= \sqrt{ 25 + 81 }$

$= \sqrt{\textbf{106}}$

Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

$(x_1, y_1) = (4, 5)$

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}$

Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$\bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2$

i.e, 53 + 53 = 106

Hence, the triangle is a right - angled triangle as well.

7 0

2 years ago

david has 212 feet of chain link fencing that he used to make a rectangular garden space in his backyard. the lengthof the garde

joja [24]

Answer:

calcium is in shells and bones

Step-by-step explanation:

4 0

2 years ago

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$

6 0

2 years ago

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