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3 years ago

Really need help Solve the system of equation Y=8x+11 Y=5x+29 X= Y=

Mathematics

1 answer:

KonstantinChe [14]3 years ago

3 0

Answer:

(6,35)

Step-by-step explanation:

Set both equations equal to each other and solve for x

8x+11=5x+29, Subtract 5x from both sides

3x+11=29, Subtract 11 from both sides

3x=18 Divide by 3

x=6

Substitute x=6 back into either equation to solve for y

y=8(6)+11

y=48+11

y=59

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Find the measure of each missing angle in each triangle. (I need the explanation as well :( )

Andrew [12]

Answer:

L= 42

M= 88

N= 50

X= 42

Y= 88

Z= 50

Step-by-step explanation:

X and L have congruent angles

M and Y have congruent angles as well

Then to find X and L do 180-50 and subtract 88 as well.

6 0

3 years ago

Somebody please help asap

Elena L [17]

Answer:

B. $4x^2 + \frac{3}{2}x - 7$

Step-by-step explanation:

$f(x) = \frac{x}{2} - 3$

$g(x) = 4x^2 + x - 4$

(f + g)(x) = f(x) - g(x)

= $\frac{x}{2} - 3 + 4x^2 + x - 4$

Add like terms

$= 4x^2 + \frac{x}{2} + x - 3 - 4$

$= 4x^2 + \frac{3x}{2} - 7$

$= 4x^2 + \frac{3}{2}x - 7$

6 0

3 years ago

Pls help no need for long explanations just a summary

mr Goodwill [35]

Answer:

FG - it is midline. FG = 5.

Explanation:

A midline in a triangle is a line segment connecting the midpoints of two sides.

A midline is half the length of the side to which it is parallel.

$\dfrac{CE}{FG} =2\\\\ \dfrac{9x-26}{7x-23} =2\\\\9x-26=2(7x-23)\\9x-26=14x-46\\14x-9x=46-26\\5x=20\\x=4$

CE = 9 · 4 - 26 = 10

FG = 10/2 = 5

8 0

3 years ago

Write an equation to represent the following statement. 24 is 4 times as great as k.

Artyom0805 [142]

Answer:

24 x 4 = k x 4

Step-by-step explanation:

8 0

3 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago