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Radda [10]
4 years ago
14

Which is an equation of the line with a slope of (1,4) and a y-intercept of-2​

Mathematics
1 answer:
anzhelika [568]4 years ago
7 0

Answer:

y=6x-2

Step-by-step explanation:

Fisrt things first, you want to find the slop. One of the points is (1,4). You want the other point. Using the y-int, you can figure out the other point is (0,-2). You get this because the y-intercept means the x value is 0.

Next you use the two points and plug it into the formula (change in y/ change in x)

(-2-4)/(0-1)=-6/-1 = 6

Now you know the slope is 6. Since you know the y-int, you can plug that into the slope-intercept formula (y=mx+b)

That would give you: y=6x-2

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65 is 52% of what number
svp [43]
65 is 52% of 125.

I do the IS over OF operation when finding a percentage, where you multiply the is by 100 then divide by the percentage. 


6 0
3 years ago
The heights of 20-year-old females are normally distributed with a mean of 64 inches 
Oduvanchick [21]

Answer:

1.5

Step-by-step explanation:

Applying,

Z = (x-μ)/σ................. Equation 1

Where Z = Z score for a height of 67 inches, x = height, μ = mean of the height, σ = standard deviation of the height.

From the question,

Given: x = 67 inches, μ = 64 inches, σ = 2 inches.

Substitute these values into equation 1

Z = (67-64)/2

Z = 3/2

Z = 1.5

6 0
3 years ago
What is the product of the two binomials below (3A+4B)(3A-4B)
Mariulka [41]

Answer:

9A^2 - 16B^2.

Step-by-step explanation:

(3A+4B)(3A-4B) = 9A^2 + 12AB - 12AB - 16B^2 = 9A^2 - 16B^2.

Hope this helps!

5 0
3 years ago
1. Find the vertices and locate the foci for the hyperbola whose equation is given.
Irina18 [472]
\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
 vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\
\textit{now let's take a look at yours}
\\\\\\
49x2 - 16y2 = 784\implies \cfrac{49x^2}{784}-\cfrac{16y^2}{784}=1
\\\\\\
\cfrac{x^2}{16}-\cfrac{y^2}{49}=1\implies \cfrac{(x-0)^2}{4^2}-\cfrac{(y-0)^2}{7^2}=1
\\\\\\
recall\implies center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})
\\\\\\


\bf \textit{now, for the foci, the foci are "c" distance from the center point}\\\\\
whereas\qquad c=\sqrt{a^2+b^2}\qquad \textit{ that is }\qquad  h\pm \sqrt{a^2+b^2}

notice your "a" and "b" components, to get the distance "c" from the center to either foci and the vertices, of course, are h + a, k and h - a, k
5 0
3 years ago
If f(x)=ax+b and f(2)=1 and f(-3)=11, what is the value of A?
allochka39001 [22]

Answer:

a = -2

Step-by-step explanation:

f(x)=ax+b

f(2)=1  

f(-3)=11

f(2) = 1   means   2a+b =1

f(-3)=11   means   -3a + b = 11

Subtracting the two equations

-(-3a +b =11) becomes 3a -b = -11   so we can add

2a+b =1

3a - b = -11

----------------------

5a = -10

Divide by 5

5a/5 = -10/5

a=-2

7 0
4 years ago
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