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Mila [183]
3 years ago
14

Calculate the angle between the longest side and the diagonal of a 577mm by 1000mm rectangle

Mathematics
1 answer:
krok68 [10]3 years ago
8 0

This should help in the photo

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Geometry! Please help!
Travka [436]
The answer is 315 as the formula is : a+b/2•h.
3 0
3 years ago
Read 2 more answers
How do I find the volume?​
kotykmax [81]

Answer:

14ft

First get the cones volume formula v=(3.14*r^2*h)/3

Second find the radius which is 1/2 the diameter.

Third plug in given and solve 366= (3.14*5^2*h)/3

366*3=3*(3.14*25*h)/3

1098=78.5h

1098/78.5=78.5h/78.5

13.987=h

5 0
3 years ago
Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
GaryK [48]

Consider all parabolas:

1.

y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0
3 years ago
2^x + 2^-x =5/2 pls I need d. ans​
yKpoI14uk [10]

Answer:

x= 1, -1

Step-by-step explanation:

6 0
3 years ago
The lines are parallel. Find the slope-intercept form of the equation y₂.
Bas_tet [7]
Since the lines are parallel and you know the slope of y_{1} is -2, you know the slope of line y_{2} will also be -2. You can see on the coordinate grid that line y_{2} crosses the y-axis at -1. When you piece all of these facts together into slope-intercept form you get y_{2} = -2x - 1.
6 0
3 years ago
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