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Finger [1]
3 years ago
7

What is the equation of the line in slope-intercept form? y = _ x+_=

Mathematics
1 answer:
pashok25 [27]3 years ago
4 0

Answer:

y=3/5x+3

Step-by-step explanation:

So find 2 points, in this case it will be

(-5,0) and (0,3)

Use the slope formula---> y2-y1=x2-x1 and enter the points above

= 3-0/0-(-5)

3/5

Slope= 3/5

y=3/5x+b

then enter (0,3)

3=0+b

b=3

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How do you graph y=-3x-5 y=x+9
irga5000 [103]
I used Desmos to graph it but here are the coordinates used to graph it in the pictures below.

3 0
3 years ago
To most closely estimate the difference below, would you round the numbers to the nearest ten thousand, the nearest thousand, or
zhannawk [14.2K]

Answer:

To most closely estimate the difference, I would round 62,980 to the nearest thousand, so it will be 63,000. That is because when rounding, 63,000 is large enough so that adding will be easy, and 62,980 is relatively close to 63,000.

I would round 49,625 to the nearest thousand, since it would make the subtraction easier and 49,625 is only 375 units away from 50,000.

63,000 - 50,000 = about 13,000.

Hope this helps!

6 0
3 years ago
Read 2 more answers
Kyla makes a triangular school pennant. The area of the triangle is 180 square inches. The base of the pennant is z inches long.
kykrilka [37]
We can represent the base as z and the height as 2z+6. We are going to use the formula A=1/2*b*h and solve for z
180=1/2*z*(2z+6)
360=2z^2+6z 
0=2z^2+6z-360
0=2(z^2+3z-180)
0=(z+15)(z-12)
So z=-15 and 12 but it must be positive so then the base is equal to 12

When we plug this into 2z+6 we get 30 for the height
2(12)+6=30

Hope this helps
4 0
3 years ago
A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
3 years ago
Urgent. Factor the trinomial by grouping. Please show work. 14x^2y+9xy^2+y^3. If prime, show how you got prime answer.
ch4aika [34]

Answer:

y x (2x +y) x (7x +y)

Step-by-step explanation:

14x^2y + 9xy^2 +y^3

(Factor the expression)

y x (14x^2 +9xy +y^2)

(Rewrite the expression)

y x (14x^2 + 7xy +2xy +y^2)

(Factor the expression)

y x (7x x(2x +y) + y x (2x +y) )

(Factor the expression)

y x (2x +y) x (7x +y)

:))

3 0
3 years ago
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