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3 years ago

Help asap plz fast just need awncers

Mathematics

1 answer:

frutty [35]3 years ago

6 0

I only know the answers for 10 and 11

10. 903.6 divided by 25.1 = 36 Estimate: 36
11. 760.38 divided by 0.38 = 2,001 Estimate: 2,000

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Kelly tells you that when variables are in the denominator, the equation becomes unsolvable. "There is a value for x that makes

saveliy_v [14]

The problem can be translated into an equation that is something like 4/5 + 3/x = 1/2
we cannot have x equal to zero because the number can be infinite.
So the LCD here is 10x, so multiply both sides by that to get:
8x + 30 = 5x
Subtract 5x and 30 from both sides:
3x = -30
divide:
x = -10
The solution isn't zero so there is a solution.

8 0

3 years ago

A $1500 loan has an annual interest rate of

Goshia [24]

127.50 = 1,500(0.0425) * x

127.50 = 63.75 * x

x= 127.50/63.75

x=2

4 0

3 years ago

6.92 as a mixed number in simplest form

tamaranim1 [39]

173/25 is the answer to that.

5 0

3 years ago

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

↓ ↓

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

Here's some points .

mars1129 [50]

Answer:

thanks mate

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers