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3 years ago

Ax + by = c Solve for b

Mathematics

1 answer:

klio [65]3 years ago

3 0

Answer:

$b = \frac{c - ax}{y}$

Step-by-step explanation:

$ax + by = c \\ by = c - ax \\ b = \frac{c - ax}{y}$

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4 years ago

A recipe requires 5/6 of a cup of sugar. If mrs.Marina is going to make one half of the recipe, how much sugar does she need

NemiM [27]

Answer:

$Half\ of\ the\ recipe\ requires\ \frac{5}{12}\ of\ a\ cup\ of\ sugar.$

Step-by-step explanation:

As given

$A\ recipe\ requires\ \frac{5}{6}\ of\ a\ cup\ of\ sugar.$

i.e

$\frac{5}{6}\ cup\ of\ sugar = 1\ recipe$

As the mrs.Marina is going to make one half of the recipe .

$Thus\ multiply\ above\ by\ \frac{1}{2}.$

We get

$\frac{5\times 1}{6\times 2}\ cup\ of\ sugar = \frac{1}{2}\ recipe$

solving the above

$\frac{5}{12}\ cup\ of\ sugar = \frac{1}{2}\ recipe$

$Therefore\ half\ of\ the\ recipe\ requires\ \frac{5}{12}\ of\ a\ cup\ of\ sugar.$

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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