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3 years ago

Rounded 11,062,113 to the nearest million

Mathematics

2 answers:

Firlakuza [10]3 years ago

7 0

Answer:

12,000,000

Step-by-step explanation:

12,000,000 according to the simply rule of mathmatics

Mashcka [7]3 years ago

4 0

Answer:

Hi there!

Your answer is;

11million

Step-by-step explanation:

In this equation, the million is in position 1<u>1</u>,062,113

We round down, because the following number is 0!

Hope this helps

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A tank having length, breadth and height 10 m, 9 m

Ne4ueva [31]

Answer:

length times width times height

Step-by-step explanation:

width is height by the way

so 20×10×9 gives you 1800

1800 petrol can hold on Meters

7 0

3 years ago

60 is 40 percent of what number

nydimaria [60]

240 plz like my awnser it would mean alot

6 0

3 years ago

An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the radius, in inche

nikklg [1K]

Wow this is a doozy! First you have to figure out what is it you are looking for? If you make a dot in the center of the triangle (which is also the center of the circle) and draw a line from the center to one of the vertices of the triangle you have the radius of the triangle and also of the circle. If you draw all 3 radii from the triangle's center to its vertices, you see you have created 3 triangles within that one triangle. The trick here is to figure out what your triangle measures are as far as angles go. If we take the interior measures of those 3 triangles, we get that each one has a measure of 120 (360/3=120). So that's one of your angles, the one across from the side measuring 6. Because of the Isosceles Triangle theorem, we know that the 2 base angles have the same measure because the sides are the same. Subtracting 120 from 180 gives you 60 which, divided in half, makes each of those remaining angles measure 30 degrees. So if we extract that one triangle from the big one, we have a triangle with angles that measure 30-30-120, with the base measuring 6 and each of the other sides measuring 5. If we then split that triangle into 2 right triangles, we have one right triangle with measures 30-60-90. Dropping that altitude to create 2 right triangles not only split the 120 degree angle at the top in half, it also split the base side of 6 in half. So our right triangle has a base of 3 and we are looking for the hypotenuse of that right triangle. WE have to use right triangle trig for that. Since we have the top angle of 60 and the base of 3, we can use sin60=3/x. Solving for x we have x=3/sin60 which gives us an x value of 3.5 inches rounded from 3.464. I'm not sure what you mean by a mixed number unless you mean a decimal, but that's the radius of that circle.

6 0

3 years ago

Please help!! MathsWatch, Surd Expressions, Clip 207c

dexar [7]

Answer:

$A=(72+36\sqrt{3})\ mm^2$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

Let

a ----> the height of rectangle in mm

b ---> the base of rectangle in mm

step 1

Find the base of rectangle

$b=AB+BC+CD+DE$ ----> by segment addition postulate

substitute

$b=3+3+3+3=12\ mm$

step 2

Find the height of rectangle

$a=FB+CG+GH$ ---> by segment addition postulate

substitute the given values

$a=3+CG+3\\a=6+CG$

Find the length sides CG

Applying the Pythagorean Theorem

$BG^2=BC^2+CG^2$

substitute the given values

$6^2=3^2+CG^2$

$CG^2=6^2-3^2$

$CG=\sqrt{27}\ mm$

simplify

$CG=3\sqrt{3}\ mm$

therefore

$a=(6+3\sqrt{3})\ mm$

step 3

Find the area of rectangle

$A=ab$

we have

$a=(6+3\sqrt{3})\ mm$

$b=12\ mm$

substitute

$A=(6+3\sqrt{3})(12)=(72+36\sqrt{3})\ mm^2$

8 0

3 years ago

1. The sum of the three consecutive members of the geometric sequence is 13. The proportion of the third and the first member is

aev [14]

If a = first term and r = common ratio we have

a + ar + ar^2 = 13 and ar^2 / a = r^2 = 9

so r = 3

and a + 3a + 9a = 13
so a = 1

so they are 1,3 and 9

2.
in geometric series we have

4 , 4r ,4r^2 , 60

Arithmetic;
4, 4r , 4r + d , 4r + 2d

so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30
so d = 30 - 2r
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0

2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24

and the numbers are 4*3 = 12 , 4*3^2 = 36

first 3 are 4 , 12 and 36 ( in geometric)
and last 3 are 12, 36 and 60 ( in arithmetic)

The 2 numbers we ause are 12 and 36.

7 0

3 years ago