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kaheart [24]
3 years ago
13

What is the slope of 12x + 3y =7

Mathematics
2 answers:
Tresset [83]3 years ago
6 0
X=7/12-y/4






Ik this because I did it
ahrayia [7]3 years ago
4 0

Answer: y = -4x + 7/3 so the slope is -4

Step-by-step explanation:

First, we need to put the linear equation into the slope-intercept form: y = mx + b

We do this by subtracting 12x from both sides, so the equation now looks like 3y = -12x + 7. Now we divide both sides by 3 to get y alone, which gives us y = -4x + 7/3

m is always the slope, so the slope of this line is -4 or -4/1.

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Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
2 years ago
select yes or no to indicate if a rectangle with the given dimensions whould have a perimeter of 50 inches
umka21 [38]
1a) no it would be 54

1b) yes

1c) yes

1d) no it would be 40

1e) yes



hope this helped
8 0
3 years ago
When multiplying 4.73 by 2.1 how many decimal places will the product have
hichkok12 [17]

it would be 9.93300

so 3 decimal places because the zeros don't count


3 0
3 years ago
Read 2 more answers
Justin is constructing a line through point Q that is perpendicular to line n. He has already constructed the arcs shown. He pla
krek1111 [17]
<span>B. It must be the same as when he constructed the arc centered at point A. This problem would be a lot easier if you had actually supplied the diagram with the "arcs shown". But thankfully, with a few assumptions, the solution can be determined. Usually when constructing a perpendicular to a line through a specified point, you first use a compass centered on the point to strike a couple of arcs on the line on both sides of the point, so that you define two points that are equal distance from the desired intersection point for the perpendicular. Then you increase the radius of the compass and using that setting, construct an arc above the line passing through the area that the perpendicular will go. And you repeat that using the same compass settings on the second arc constructed. This will define a point such that you'll create two right triangles that are reflections of each other. With that in mind, let's look closely at your problem to deduce the information that's missing. "... places his compass on point B ..." Since he's not placing the compass on point Q, that would imply that the two points on the line have already been constructed and that point B is one of those 2 points. So let's look at the available choices and see what makes sense. A .It must be wider than when he constructed the arc centered at point A. Not good. Since this implies that the arc centered on point A has been constructed, then it's a safe assumption that points A and B are the two points defined by the initial pair of arcs constructed that intersect the line and are centered around point Q. If that's the case, then the arc centered around point B must match exactly the setting used for the arc centered on point A. So this is the wrong answer. B It must be the same as when he constructed the arc centered at point A. Perfect! Look at the description of creating a perpendicular at the top of this answer. This is the correct answer. C. It must be equal to BQ. Nope. If this were the case, the newly created arc would simply pass through point Q and never intersect the arc centered on point A. So it's wrong. D.It must be equal to AB. Sorta. The setting here would work IF that's also the setting used for the arc centered on A. But that's not guaranteed in the description above and as such, this is wrong.</span>
8 0
3 years ago
Read 2 more answers
Calculus question....does anyone know how to solve this?
iogann1982 [59]
h(x)=(f\circ g)(x)
\sqrt{x+5}=\sqrt{g(x)+2}
x+5=g(x)+2
g(x)=x+3

Just to check this is correct:

(f\circ g)(x)=f(g(x))=f(x+3)=\sqrt{(x+3)+2}=\sqrt{x+5}=h(x)
6 0
3 years ago
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