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2 years ago

What’s the slope and y - intercept of this equation ?

Mathematics

2 answers:

Jlenok [28]2 years ago

8 0

Slope 4x and y-intercept is -10 I think I’m positive

Artyom0805 [142]2 years ago

5 0

Answer:

y-intercept is -10 and slope is 4

Step-by-step explanation:

in y= 4x-10 the slope is 4 because in the formula y=mx+b, m is the slope and the y-intercept is -10 because also in the formula y=mx+b, b is the y-intercept

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Can u help me with 8 and 10

Wewaii [24]

6

I GUESSING BRO I AINT GOT TIME FOR THAT B

8 0

3 years ago

Can someone please help me with these two please I'll appreciate it :)

s344n2d4d5 [400]

4.

$\sqrt{64}=8\\\\8\dfrac17=8.14285714...\\\\8.\overline{14}=8.14141414...\\\\ \dfrac{15}2=7,5\\\\\\ \boxed{\ \dfrac{15}2\ \ \textless \ \ \sqrt{64}\ \ \textless \ \ 8.\overline{14}\ \ \textless \ \ 8\dfrac17\ }$

5.

$\sqrt{18}$ doesn't belong with the other three because it is irrational number
The other three can be expressed as a ratio between two integers.

$-\dfrac{10}2=\dfrac{-10\ }2\\\\-13.4=-13\frac4{10}=\frac{-134}{\ 10}\\\\ 22.\overline7=22\frac79=\frac{205}9\\\\\\ \sqrt{18}=\sqrt{9\cdot2}=\sqrt9\cdot\sqrt2=3\sqrt2$

7 0

4 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

Find the area bounded by y=ln x, x=2, x=4 and the x-axis.

Advocard [28]

4 sq.units

area of a rectangle = length × breadth

length = 4units

breadth = 1units

A = l × b

= 4 × 1

= 4 sq.units

<h3>HOPE IT HELPS U </h3>

IF YES 

MARK ME AS THE BRAINLIEST

8 0

3 years ago

How far did we jump to go from 17 to 20 r=26-17 17+r=26

BabaBlast [244]

Answer:

Both are correct!

Step-by-step explanation:

if R = 26 - 17 = 9

now the next one

17 + R = 26

Same thing as R = 26 - 17 = 9

5 0

3 years ago

Read 2 more answers