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2 years ago

X/-3+33=30 solveeeeee

Mathematics

1 answer:

Montano1993 [528]2 years ago

6 0

Answer:

x = 9

Step-by-step explanation:

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faltersainse [42]

Answer:

D.mode

Step-by-step explanation:

4 0

3 years ago

Question 5 of 10

Lady_Fox [76]

Answer:

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x = − 0. 142857

Step-by-step explanation:

7 0

2 years ago

A bookkeeper has a balance of $83.92. If she collects $76.75

kotegsom [21]

Answer:

The bookkeeper already has a balance of 83.92. She collects 76.75 so that's 83.92+76.75 which equals to 160.67. Then she pays out 38.58 so that's a deduction to what she already has. 160.67-38.58=122.09.

SO the bookkeeper has $122.09 left.

5 0

3 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

3 years ago

Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$

Thus the solution is found

5 0

3 years ago