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Stella [2.4K]
3 years ago
15

How do you find how many times larger

Mathematics
1 answer:
masya89 [10]3 years ago
4 0

Answer

whats the problem.

Step-by-step explanation:

 

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Which graph shape represents the polar curve r = 1 + 4sin(θ)?
umka2103 [35]

Answer:

A cardioid

Step-by-step explanation:

A cardioid is a curve in the shape of a heart traced by a point on the circumference of a circle as it rolls around a fixed circle of the same radius.

The equation of vertical cardioid is r=a ± a\sin \theta

Given the polar curve is r= 1+ 4 \sin \theta

This is an equation of vertical cardioid.

So, it represents a cardioid.

7 0
3 years ago
Read 2 more answers
My daughter is in 3rd grade. On a 3rd grade level how can she explain that 5/6 is greater than 5/8
blagie [28]
Well,5/6 is closer to one than 5/8 is,just add 1/6+5/6 and then you get 1,adding 1/8 to 5/8 is only 6/8,so that's how you can tell how it is greater! Hope this helps your daughter!
4 0
3 years ago
Read 2 more answers
Heeeeeeeeeeeeeeeeeeepl
DIA [1.3K]

Answer:

Hi there!

Your answer is;

a)

i) 400% of 240

240 is 100%

× 4

960= 400%

ii) 40% of 240

100% = 240

/100

1% = 2.4

× 40

40% = 96

iii) 4% of 240

100% is 240

/100

1% = 2.4

× 4

4% = 9.6

iv) .04% of 240

100% = 240

/100

1% = 2.4

/100

.01% = .024

× 4

.04% = .096

b) the patterns is that all these numbers equal sometime 96. Each of these have a different decimal place, but have the same actual numbers.

c) 4000% = 240

take the pattern:

400% is 960

Scale it up to 4000 by 10

400% is 960

× 10

4000% is 9600

Hope this helps!

5 0
3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term 2x^2, which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form  f(x)=ax^2+bx+c is given by the expression:

x_v=\frac{-b}{2a}

Since in our case a=2 and b=-3, we get that the x-position of the vertex is:

x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6

See the graph produced in the attached image.

4 0
3 years ago
I have no idea help please
Sliva [168]

Answer:

C

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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