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2 years ago

(5,-7) (5,3) finding slope using formula.

Mathematics

1 answer:

aleksley [76]2 years ago

6 0

m = (y2 -y1)/ (x2-x1)

m = (3-(-7))

--------

5-5

The slope is undefined since the denominator is zero

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Solve by factorising<br><br><img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="abs

Paladinen [302]

Step-by-step explanation:

$\underline{ \underline{ \text{Solving \: a \: quadratic \: equation \: by \: factorisation \: method}}} :$

In this method , the second order of polynomial ax² + bx + c is factorised and expressed as the product of two linear factors. Then each linear factor is separately solved to get the required solutions of the equation by applying zero factor property. In zero factor property , if p•q = 0 , then either p = 0 or q = 0 . In other words , if the product of two numbers is 0 , then one or both of the numbers must be 0.

$\underline{ \underline{ \text{Let's \: solve}}} :$

$\tt{ {x}^{2} + 16x + 55 = 0}$

Here , we have to find the two numbers that multiplies to 55 and adds to 16.

⤑ $\tt{ {x}^{2} + (11 + 5)x + 55 = 0}$

⤑ $\tt{ {x}^{2} + 11x + 5x + 55 = 0}$

⤑ $\tt{x(x + 11) + 5(x + 11) = 0}$

⤑ $\tt{(x + 5)(x + 11) = 0}$

Either :

$\tt{x + 5 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: Or : \: \: x + 11 = 0}$

$\sf{⟶ x = 0 - 5} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⟶ x = 0 - 11$

$\tt{⟶x = - 5 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:⟶ x = - 11$

$\red{ \boxed{ \boxed{ { \tt{Our \: final \: answer : \boxed{ \underline{ \bold{ \tt{x = - 5 \: , - 11}}}}}}}}}$

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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6 0

2 years ago

Functions! Help with number 3 please!!!

Norma-Jean [14]

Answer: 20

Step-by-step explanation:

Assuming g(x) = -4x + 1,

( -4 (-4) + 1) - (-4 + 1)

= (16 + 1) - (-3)
= 17 + 3

= 20

3 0

1 year ago

An oil-well contractor drills a shaft 7 meters deeper into the ground every 2 hours. Which graph has a slope that best represent

pentagon [3]

Answer:

Graph G

Step-by-step explanation:

The graphs are attached

Given that the oil-well contractor drills a shaft 7 meters deeper into the ground every 2 hours, hence the rate at which the shaft drills = 7 meter / 2 hours = 3.5 meters per hour

Since the drill goes into the ground, hence the rate is negative that is -3.5 meters per secong

The slope (rate of change) of a line (m) is given by:

$m=\frac{y_2-y_1}{x_2-x_1}$

a) For graph F, the line passes through the point (0,0) and (10, -20). Hence:

$Slope\ of\ line\ F=\frac{-20-0}{10-0} =-2\ meters\ per\ second$

b) For graph G, the line passes through the point (0,0) and (10, -70). Hence:

$Slope\ of\ line\ G=\frac{-70-0}{10-0} =-7\ meters\ per\ second$

c) For graph H, the line passes through the point (0,0) and (10, -35). Hence:

$Slope\ of\ line\ H=\frac{-35-0}{10-0} =-3.5\ meters\ per\ second$

d) For graph J, the line passes through the point (0,-3.5) and (10, -3.5). Hence:

$Slope\ of\ line\ j=\frac{-3.5-(-3.5)}{10-0} =0\ meters\ per\ second$

8 0

2 years ago

Read 2 more answers

What kind of financial institutions does the Office of Thrift Supervision oversee?

xxTIMURxx [149]

Savings and Loan Institutions.

5 0

2 years ago

What is the area of the figure shown?

musickatia [10]

Check the picture below.

now, we're making an assumption that, the two blue shaded region are equal in shape, and thus if that's so, that area above the 14 is 6 and below it is also 6, 14 + 6 + 6 = 26.

so hmm if we simply get the area of the trapezoid and subtract the area of the yellow triangle and the area of the cyan triangle, what's leftover is what we didn't subtract, namely the shaded region.

$\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ h=15\\ a=14\\ b=26 \end{cases}\implies A=\cfrac{15(14+26)}{2}\implies A=300 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\stackrel{trapezoid}{300}~~ - ~~\stackrel{yellow~triangle}{\cfrac{1}{2}(26)(9)}~~ - ~~\stackrel{cyan~triangle}{\cfrac{1}{2}(15)(6)}} \\\\\\ 300~~ - ~~117~~ - ~~45\implies 138\qquad \textit{blue shaded area}$

8 0

1 year ago