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3 years ago

Find the slope-intercept form of the equation of the line that passes through

Mathematics

1 answer:

yawa3891 [41]3 years ago

5 0

Answer:

Step-by-step explanation:

12x - 3y = 10

-3y = -12x + 10

y = 4x - 10/3....the slope here is 4

parallel lines will have the same slope

y = mx + b

slope(m) = 4

(-5,3)...x = -5 and y = 3

now we sub and find b, the y int

3 = 4(-5) + b

3 = -20 + b

3 + 20 = b

23 = b

so ur equation is : y = 4x + 23 <==

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Mr. Britt is pouring a 36' x 58' concrete pad for a metal building. If concrete costs

Helga [31]

Answer:

Step-by-step explanation:

8 in × (1 ft)/(12 in) = ⅔ ft

36 ft × 58 ft × ⅔ ft = 1392 ft³

1392 ft³ × (1 yd³)/(27 ft³) ≅ 52 yd³

52 yd³ × $123/yd³ = $6396

4 0

3 years ago

A polynomial that can be factored is called

Bingel [31]

<span>reducible.

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4 0

3 years ago

Geometry pls help !!! Find the value of AB.<br> AB = [?]

V125BC [204]

Answer:

AB = 16 Units

Step-by-step explanation:

In the given figure, CD is the diameter and AB is the chord of the circle.

Since, diameter of the circle bisects the chord at right angle.

Therefore, AE = 1/2 AB

Or AB = 2AE...(1)

Let the center of the circle be given by O. Join OA.

OA = OD = 10 (Radii of same circle)

Triangle OAE is right triangle.

Now, by Pythagoras theorem:

$OA^2 = AE^2 + OE^2 \\10^2 = AE^2 + 6^2 \\100= AE^2 + 36\\100-36 = AE^2 \\64= AE^2 \\AE = \sqrt{64}\\AE = 8 \\\because AB = 2AE..[From \: equation\: (1)] \\\therefore AB = 2\times 8\\\huge \purple {\boxed {AB = 16 \: Units}}$

4 0

3 years ago

Rewrite the fraction without an exponent (7/8)^-2

Artist 52 [7]

Answer:

64/49 or 1 and 15/49

Step-by-step explanation:

1. process of exponents turns it into 8/7 ^2

2. Solve and simplify

5 0

3 years ago

Read 2 more answers

Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30

Dafna11 [192]

Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

$\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}$

$\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~ \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------$

$\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o$

8 0

2 years ago