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3 years ago

How do I solve this

Mathematics

1 answer:

MatroZZZ [7]3 years ago

4 0

Answer:

Step-by-step explanation:

First do -1 - 5, which is -6.

then square -6.

so 6 x 6 equals 36.

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WILL GIVE BRAINLIEST

puteri [66]

Step-by-step explanation:

${EF}^{2} = {EG}^{2} - {FG}^{2} \\ \\ = {8}^{2} - {6}^{2} \\ \\ = 64 - 36 \\ \\ = 28 \\ \\ \therefore \: {EF} = \sqrt{28} \\ \\ \therefore \: {EF} = 2\sqrt{7} \\ \\ \sin \: E = \frac{FG }{EG} = \frac{6}{8} = \frac{3}{4} \\ \\ \huge\red {\boxed{\therefore \: \sin \: E = \frac{3}{4}} } \\\sec \: G = \frac{ EG}{FG} = \frac{8}{6} = \frac{4}{3} \\ \\ \huge\purple {\boxed{ \therefore \: \sec \: G = \frac{4}{3}}} \\\\\cot \: G = \frac{ FG}{EF} = \frac{6}{2\sqrt 7} = \frac{3}{\sqrt 7} \\ \\ = \frac{3\times\sqrt 7 }{\sqrt 7\times\sqrt 7 } \\ \\ = \frac{3\sqrt 7 }{ 7 } \\ \\ \huge\orange {\boxed{ \therefore \: \cot \: G = \frac{3\sqrt 7 }{ 7 } }} \\\\$

6 0

3 years ago

What property would you use to solve. X/5 = 3

OLEGan [10]

Answer:

Multiplication

Step-by-step explanation:

x/5 = 3

Multiply by 5

x = 15

5 0

3 years ago

Joaquin writes the following list of numbers.

swat32

From the given list of numbers, the numbers that are:

rational are 26, -3/2, 0, and 9.

irrational are 5.737737773..., and √45.

Any number that can be written in the form of p/q, where p and q are integers, and q ≠ 0, are called rational numbers.

All terminating and non-terminating recurring decimals are rational numbers.

All the numbers that cannot be represented in the rational form of p/q are irrational numbers.

All non-terminating non-recurring decimals are irrational.

All square roots of imperfect square numbers, that is, surds, are irrational numbers.

In the question, we are asked to classify the given list of numbers into rational and irrational numbers.

5.737737773...: It is an irrational number, since its a non-terminating non-recurring decimal.
26: It is rational as it can be represented in the p/q form (26/1).
√45: It is irrational as it is a square root of an imperfect square number, that is, it is a surd.
-3/2: It is rational as it is in the form p/q.
0: It is rational as it can be represented in the p/q form (0/1).
9: It is rational as it can be represented in the p/q form (9/1).

Thus, from the given list of numbers, the numbers that are:

rational are 26, -3/2, 0, and 9.

irrational are 5.737737773..., and √45.

Learn more about rational and irrational numbers at

brainly.com/question/14994517

#SPJ9

The provided question is incorrect. The correct question is:

Joaquin writes the following list of numbers.

5.737737773..., 26, √45, -3/2, 0, 9.

Which numbers are rational?

Which numbers are irrational?

6 0

2 years ago

16x − 7 ≤ −71 find x

My name is Ann [436]

16x-7≤-71
Add 7 to both sides
16x≤-64
Divide both sides by 16
x≤ -4

Hope this helps! :)

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago