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Doss [256]
4 years ago
12

Use mental math to complete the pattern5×6

Mathematics
1 answer:
7nadin3 [17]4 years ago
6 0
5x6 = 30

5, 10, 15, 20, 25, 30, etc

1, 2 , 3, 4, 5, 6
You might be interested in
The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.
Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m  + 40 m  + c = 120 m

61 m +  c = 120 m

c = 120 m - 61 m

c = 59 m

                         B

                     ↓            ↓  

                  ↓                          ↓

                ↓                                       ↓

            A →  →  → →  →  → →  → →    →    →  C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0

Angle B is calculated as follows;

Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0

Angle C is calculated as follows;

Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0

Therefore, the largest angle of the field is 149⁰.

       

8 0
3 years ago
An Xbox that normally sells for $390 is on sale at a 30% discount. What is the sale price of the Xbox?
schepotkina [342]
$273 as 390×(1-30%)=273
8 0
3 years ago
Use the following equations to solve the problem.
Liono4ka [1.6K]
35 mph = 56 km/h

Final distance = 11.3 + 56 x 3
Final distance = 179.3 km

The second option is correct.
3 0
3 years ago
Read 2 more answers
Acellus. please help me
Doss [256]

by-step explaAnswer:

by-step epla

Step-nation:

a

6 0
3 years ago
The coordinate plane below represents a city. Points A through F are schools in the city.
Usimov [2.4K]
Part A. The technique on how to find the equation that only applies to point D and E, is to create a line or curve that only includes two of these points. In this case, I created a random parabola that isolates points C and F from the rest of the points. First, we have to find the equation of the parabola through its general forms:

(x - h)² = +/-4a(y-k)      or    (y - k)² = +/-4a(x - h)

For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:

(-2 - 0)² = +4a(4 - 3)
4a = 4

So, the equation of the parabola is:

x² = 4(y - 3)
x² = 4y - 12

Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4

Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).

3 ? 7(4) − 4
3 ? 24
3 < 24

Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F. 

5 0
3 years ago
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