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sineoko [7]
3 years ago
9

Plz help befor it’s late my parents get mad so help me please

Mathematics
1 answer:
lapo4ka [179]3 years ago
3 0
The answer to 8 is 6 the answer to 9 is 248 the answer to 10 is 11 days and 11 is 14
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What is the answer to solve -2=4r+s for s
sleet_krkn [62]
The answer is s= -(4r+2)
7 0
3 years ago
If I have $1,050 dollars, and I want to buy something 6 times for $45 dollars, how much times can I buy it until I run out of mo
joja [24]

Answer:

140

Step-by-step explanation:

here,

You have 1050

now

$1050(45×6)

=140 items

8 0
2 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:

<em>(a) x=2, y=-1</em>

<em>(b)  x=2, y=2</em>

<em>(c)</em> \displaystyle x=\frac{5}{2}, y=\frac{5}{4}

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

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

(c) The system to solve is

\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
4 years ago
PLEASE PLEASE HELP!!!!! WILL GIVE BRAINLIEST!!!!!
fiasKO [112]

To be written in standard form, the exponents need to be in order from largest to smallest.

The answer is: a. −3x^5 + 4x^3 + 10x^2

6 0
3 years ago
If 2tanA=3tanB,then prove that tan(A-B)=sin2B/(5-cos2B) ...?
andreev551 [17]
Tan ( A - B ) = ( tan A - tan B ) / ( 1 + tan A tan B )
tan A = 3 tan B/2
tan ( A - B ) = ((3 tan B/ 2)-tan B) / ( 1 + 3 tan² B/2)=
= (tan B/2)  / ( 2 + 3 sin²B/cos²B )=
= (sin B / cos B) / (( 2cos² B+3sin²B)/cos²B)=
=( sin B cos B ) / ( 2 cos²B + 3 ( 1 - cos² B ) ) =
= (sin B cos B ) / ( 2 cos² B + 3 - 3 cos² B ) =
= ( sin 2 B ) / 2 ( 3 - cos² B ) =
= ( sin 2 B ) / ( 6 - cos² B )=
= ( sin 2 B ) / ( 5 + 1 - 2 cos² B )=
= ( sin 2 B ) / ( 5 + sin² B + cos ² B - 2 cos² B ) =
= ( sin 2 B ) / ( 5 - ( cos² B - sin² B ) ) =
= ( sin 2 B ) / ( 5 - cos 2 B )  - correct 
8 0
3 years ago
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