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3 years ago

Solve the system by substitution. y = 3x + 14 y = X

Mathematics

2 answers:

boyakko [2]3 years ago

6 0

Answer:

x = 7 and y = -7

Step-by-step explanation:

use substitution method

lys-0071 [83]3 years ago

4 0

Answer:

$\boxed{x = -7~~and~~ y =-7}$

Step-by-step explanation:

y = 3x + 14

y = x

x = 3x + 14

3x - x = -14

2x = -14

$x = \frac{-14}{2}$

x = -7

y = x

y = -7

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On a field trip there were 5 girls for every 8 boys. how many girls attended the 130-student field trip?

postnew [5]

50 girls attended the trip.
80 boy + 50 girls = 130 total students
8x10 =80
5x10=50

multiply both by 10.

4 0

3 years ago

Read 2 more answers

Given: CD is a perpendicular bisector of AB . Prove: Any point on CD is equidistant from the endpoints of AB . Match each statem

Sunny_sXe [5.5K]

Answer:

The two triangles are congruent, so any point on CD will be equidistant from endpoints of AB.

Step-by-step explanation:

Let the consider the figure as per the attached image:

AB be a line whose perpendicular bisector line is CD.

CD divides the line AB in two equal line segments making an angle of $90^\circ$ on both the sides as shown in the attached figure.

Let a point on CD be E.

Here, two triangles are formed:

$\triangle AED\ and \ \triangle BED.$

Side ED is common between the two triangles.

Also, Side ED is perpendicular bisector:

$\angle EDA = \angle EDB =90^\circ$

And Sides AD = DB

According to SAS congruence (i.e. Two sides are equal and angle between them is equal):

$\triangle ADE \cong \triangle BDE$

And as per the <em>properties of congruent triangles, all the sides are equal.</em>

$\Rightarrow EA = EB$

EA and EB is the distance of point E on line CD from the endpoints of line AB.

Hence proved that Any point on CD is equidistant from the endpoints of AB .

4 0

3 years ago

Will mark Brainlest please answer. find the value of a,b. <br>,p,q from the equal order pairs

NARA [144]

Step-by-step explanation:

<h3>Question-1:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 3p = 2p - 1 \dots \dots i\\2q - p = 1 \dots \dots ii\end{cases}$

cancel 2p from the i equation to get a certain value of p:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - p = 1 \end{cases}$

now substitute the value of p to the second equation:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - ( - 1) = 1 \end{cases}$

simplify parentheses:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q + 1= 1 \end{cases}$

cancel 1 from both sides:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q = 0\end{cases}$

divide both sides by 2:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\q = 0\end{cases}$

<h3>question-2:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\3y= x + y \dots \dots ii\end{cases}$

cancel out y from the second equation:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\ x = 2y \dots \dots ii\end{cases}$

substitute the value of x to the first equation:

$\displaystyle \begin{cases} \displaystyle 2.2y-y= 3 \\ x = 2y \end{cases}$

simplify:

$\displaystyle \begin{cases} \displaystyle 3y= 3 \\ x = 2y \end{cases}$

divide both sides by 3:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2y \end{cases}$

substitute the value of y to the second equation which yields:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2 \end{cases}$

<h3>Question-3:</h3>

by order pair we obtain;

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \dots \dots i\\3q + 2p = 3 \dots \dots ii\end{cases}$

rearrange:

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \\2p + 3q= 3 \end{cases}$

subtract and simplify

$\displaystyle \begin{array}{ccc} \displaystyle 2p + q = 2 \\2p + 3q= 3 \\ \hline - 2q = - 1 \\ q = \dfrac{1}{2} \end{array}$

substitute the value of q to the first equation:

$\displaystyle 2.p+ \frac{1}{2} = 2$

make q the subject of the equation:

$\displaystyle p = \frac{3}{4}$

hence,

$\displaystyle q = \frac{1}{2} \\ p = \frac{3}{4}$

3 0

3 years ago

Read 2 more answers

Come hither big brains

katrin2010 [14]

Answer:

The answer is c

Step-by-step explanation:

7 0

3 years ago

Find the zeros of the following<br> please solve step by step<br> 1. x³+9x²+26x+24=0

Vlad1618 [11]

Answer: -4, -3, -2

Step-by-step explanation:

By inspection, we know $x=-2$ is a root.

We can thus rewrite the equation as

$(x+2)\left(\frac{x^3 + 9x^2 + 26x+24}{x+2} \right)=0\\\\(x+2)(x^2 + 7x+12)=0\\\\(x+2)(x+3)(x+4)=0\\\\x=-4, -3, -2$

3 0

2 years ago