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vlabodo [156]
3 years ago
6

Can you please help me with this question it is confusing for me.

Mathematics
1 answer:
vova2212 [387]3 years ago
5 0
75/15=5 so the answer would be 5 months
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In Aunt Melly's attics there are also spiders and ants, the total of 136 legs and 20 heads. If a spider has 8 legs and an ant ha
anzhelika [568]

Answer:

Spiders=8

Ants=12

Step-by-step explanation:

Spider(s)=8 legs

Ants(a)=6 legs

Total legs=136

Total heads=20

8s+6a=136 (1)

s+a=20 (2)

From (2)

s=20-a

Substitute 20-a into (1)

8s+6a=136

8(20-a)+6a=136

160-8a+6a=136

160-2a=136

-2a=136-160

-2a=-24

a= -24/-2

a=12

Substitute a=12 into (2)

s+a=20

s+12=20

s=20-12

s=8

3 0
4 years ago
Calculate area and perimeter​
mina [271]

Answer:

area ≈ 12.505

perimeter ≈  16.1684

Step-by-step explanation:

We are given

- the radius of the circle (and therefore area of the circle)

- the area of the triangle

We want to find

- angle AOB/AOT. We want to find this because 360/the angle gives us how many OABs fit into the circle. For example, if AOT was 30 degrees, 360/30 = 12 (there are 360 degrees in a circle, so that's where 360 comes from). The area of the circle is equal to πr² = π6² = 36π, and because AOT is 30 degrees, there are 12 equal parts of sector OAB in the circle, so 36π/12=3π would be the area of the sector. A similar conclusion can be reached from the circumference instead of the area to find the distance between A and B along the circle, and OA + AB + BO = the perimeter of the minor sector.

First, we can say that OAT is a right triangle because a tangent line is perpendicular to the line from the center to the point on the circle, so AT is perpendicular to OA. This forms two right angles, one of which is OAT

One thing that we can start to solve is AT. We know that the area of a triangle is equal to base * height /2, and the height of this triangle is AO, with the base being AT. Therefore, we can say

15 = AO * AT / 2

15 = 6 * AT / 2

15 = 3 * AT

divide both sides by 3 to isolate AT

AT = 5

Because OAT is a right triangle, we can say that the hypotenuse ² =  the sum of the squares of the two other lengths. The hypotenuse is opposite of the largest angle (in this case, the right angle, as in a right triangle, the right angle is always the largest), so it is OT in this case. The other two sides are OA and AT, so we can say that

OA² + AT² = OT²

5²+6² = OT²

25+36=61=OT²

square root both sides

OT = √61

Next, the Law of Sines states that

sinA/a = sinB/b = sinC/c with angles A, B, and C with sides a, b, and c. Corresponding sides are opposite their corresponding angles, so in this case, AT corresponds to angle AOT, OT corresponds to angle OAT, and AO corresponds to angle ATO.

We want to find angle AOT, as stated earlier, so we have

sin(OAT)/OT = sin(ATO)/OA = sin(AOT)/AT

We know the side lengths as well as OAT/sin(OAT) and want to figure out AOT/sin(AOT), so one equation that helps us get there is

sin(OAT)/OT = sin(AOT)/AT, encompassing our 3 known values and isolating the one unknown. We thus have

sin(90)/√61 = sin(AOT) /5

plug in sin(90) = 1

1/√61 = sin(AOT)/5

multiply both sides by 5 to isolate sin(AOT)

5/√61 = sin(AOT)

we can thus say that

arcsin(5/√61) = AOT ≈39.80557

As stated previously, given ∠AOT, we can find the area and perimeter of the sector. There are 360/39.80557 ≈ 9.04396 equal parts of sector OAB in the circle. The area of the circle is πr² = 36π, so 36π / 9.04396 ≈ 12.505 as the area. The circumference is equal to π * diameter = π * 2 * radius = 12 * π, and there are 9.04396 equal parts of arc AB in the circumference, so the length of arc is 12π / 9.04396 ≈ 4.1684. Add that to OA and OB (both are equal to the radius of 6, as any point from the center to a point on the circle is equal to the radius) to get 6+6 + 4.1684 = 16.1684 as the perimeter of the sector

5 0
3 years ago
In a school there are 500 students and 35 teachers. What is the ratio of students to teachers?
valkas [14]
The ratio is 500:35, or 14.28
8 0
3 years ago
Read 2 more answers
HELP. How long is the minor axis for the ellipse shown below?
kifflom [539]

Given:

The equation of ellipse is

\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1

To find:

The length of the minor axis.

Solution:

The standard form of an ellipse is

\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1      ...(i)

where, (h,k) is center, if a>b, then 2a is length of major axis and 2b is length of minor axis.

We have,

\dfrac{(x+4)^2}{25}+\dfrac{(y-1)^2}{16}=1      ...(ii)

On comparing (i) and (ii), we get

b^2=16

Taking square root on both sides.

b=\pm 4

Consider only positive value of b because length cannot be negative.

b=4

Now,

Length of minor axis = 2b

                                  = 2(4)

                                  = 8

So, the length of minor axis is 8 units.

Therefore, the correct option is B.

8 0
3 years ago
The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 101010 denarii per day to support 444
givi [52]

<u>the correct question is</u>

The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?

Let

x-------> the cost to support a legionary per day

y-------> the cost to support an archer per day

we know that

4x+4y=10 ---------> equation 1

2x+2y=5 ---------> equation 2

If you multiply equation 1 by 2

2*(2x+2y)=2*5-----------> 4x+4y=10

so

equation 1 and equation 2 are the same

The system has infinite solutions-------> Is a consistent dependent system

therefore

<u>the answer is</u>

We cannot solve for a unique cost for each soldier, because there are infinite solutions.

5 0
3 years ago
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