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2 years ago

4x+5y=20 in point slope form

Mathematics

2 answers:

saul85 [17]2 years ago

8 0

Answer:

y = $-\frac{4}{5}$ x + 4

Step-by-step explanation:

Ax + By = C is standard form of linear equation.

y - $y_{1}$ = m( x - $x_{1}$ )

y = mx + b is slope intercept form of linear equation.

Thus, to write the point-slope equation we have to have the coordinates of the point that given line is passing through!

I assume, that we have to re-write the standard form of linear equation into slope-intercept form. That's mean, we have to solve given equation for "y".

5y = - 4x + 20

y =  $-\frac{4}{5}$  x + 4

Anna35 [415]2 years ago

4 0

Answer:

y = -4/5x - 4

Step-by-step explanation:

y - y1 = m(x-x1)

4x + 5y = 20 ( subtract 4x on both sides)

5y = -4x + 20 (divide by 5 on each side)

y = -4/5x + 4

So we know one point is (0,4) y-intercept

now plug in (0,4)

y - 4 = -4/5(x-0) (Distribute -4/5 to x - 0)

y - 4 = -4/5x (now add 4 on each side)

y = -4/5x - 4

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Write the quadratic function in vertex form. y = x2 - 2x + 5

Jlenok [28]

Vertex form formula: y = a(x-h)^2 +k, with vertex (h,k)
There are multiple ways to find the vertex. One way is to find the roots and then find the x value exactly in between them, because this parabola is symmetrical.

0 = (x - 3)(x + 2), so x = 3 and -2. The point directly in the middle is x = 1/2 = h

To find the y value of the vertex, plug in 1/2 to the equation.

(1/2)^2 - 2(1/2) + 5 = 4.25 = k

y = (x - 0.5)^2 + 4.25

4 0

3 years ago

Read 2 more answers

Need help! Does anyone know how to do this type of math? (It's geometry)

mart [117]

Answer:

Step-by-step explanation:

refer the attachment

to solve the question we need to recall one of the most important theorem of circle known as two tangent theorem which states that tangents which meet at the same point are equal that is being said

FA=AI=17
CH=CG=2.5

since $\rm BA=FA+FB$ and it's given that FA and BA are 17 and 29 FB should be

$\rm 29=17+FB$

therefore,

$FB=29-17=\boxed{12}$

once again by two tangent theorem we acquire:

$FB=BH=\boxed{12}$

As BC=BH+CH,BC is

12+2.5
$\boxed{14.5}$

likewise,AD=AI+DI so,

21=17+DI [AD=21(given) and AI=17 (by the theorem)]

thus,

DI=21-17= $\boxed{4}$

By the theorem we obtain:

DI=DG=4

Similarly,DC=DG+CH therefore,

DC=4+2.5= $\boxed{6.5}$

Now finding the Perimeter of ABCD

$P_{\text{ABCD}}=\text{AB+AD+BC+DC}$

substitute what we have and got

$P_{\text{ABCD}}=\text{29+21+14.5+6.5}$

simplify addition:

$P_{\text{ABCD}}=\boxed{71}$

hence,

the Perimeter of ABCD is 71

4 0

2 years ago

Find a set of parametric equations for y= 5x + 11, given the parameter t= 2 – x

Sati [7]

Answer:

$x = 2-t$ and $y = -5\cdot t +21$

Step-by-step explanation:

Given that $y = 5\cdot x + 11$ and $t = 2-x$ , the parametric equations are obtained by algebraic means:

1) $t = 2-x$ Given

2) $y = 5\cdot x +11$ Given

3) $y = 5\cdot (x\cdot 1)+11$ Associative and modulative properties

4) $y = 5\cdot \left[(-1)^{-1} \cdot (-1)\right]\cdot x +11$ Existence of multiplicative inverse/Commutative property

5) $y = [5\cdot (-1)^{-1}]\cdot [(-1)\cdot x]+11$ Associative property

6) $y = -5\cdot (-x)+11$ $\frac{a}{-b} = -\frac{a}{b}$ / $(-1)\cdot a = -a$

7) $y = -5\cdot (-x+0)+11$ Modulative property

8) $y = -5\cdot [-x + 2 + (-2)]+11$ Existence of additive inverse

9) $y = -5 \cdot [(2-x)+(-2)]+11$ Associative and commutative properties

10) $y = (-5)\cdot (2-x) + (-5)\cdot (-2) +11$ Distributive property

11) $y = (-5)\cdot (2-x) +21$ $(-a)\cdot (-b) = a\cdot b$

12) $y = (-5)\cdot t +21$ By 1)

13) $y = -5\cdot t +21$ $(-a)\cdot b = -a \cdot b$ /Result

14) $t+x = (2-x)+x$ Compatibility with addition

15) $t +(-t) +x = (2-x)+x +(-t)$ Compatibility with addition

16) $[t+(-t)]+x= 2 + [x+(-x)]+(-t)$ Associative property

17) $0+x = (2 + 0) +(-t)$ Associative property

18) $x = 2-t$ Associative and commutative properties/Definition of subtraction/Result

In consequence, the right answer is $x = 2-t$ and $y = -5\cdot t +21$ .

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3 years ago

Quadrilateral jklm is a rhombus. the diagonals intersect at n. if the measure of angle jkl is 104°, find the measure of angle jk

Paul [167]

The diagram of rhombus JKLM is shown in the diagram below

A rhombus is a quadrilateral with four equal sides and its diagonals intersect perpendicular to each other (makes 90° angles). Opposite angles are equal (the same with a parallelogram). Each diagonal bisects the angle at J, K, L, and M equally

If angle JKL is 104°, the measurement of angle JKN is 104÷2=51°