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3 years ago

How to work it out ‍

Mathematics

2 answers:

ELEN [110]3 years ago

7 0

Answer: b = -1

<h3>Method:</h3>

1) Expand out 2(6b + 5)

2 × 6b + 2 × 5 = 2 + 4b

12b + 10 = 4b + 2

2) Subtract 10 from both sides

12b + 10 - 10 = 4b + 2 - 10

12b = 4b - 8

3) Subtract 4b from both sides

12b - 4b = 4b - 4b - 8

8b = -8

4) Divide both sides by 8

(8b) ÷ 8 = -8 ÷ 8

b = -1

5) Substitute back in to check

2(6(-1) + 5) = 2 + 4(-1)

2(-6 + 5) = 2 + -4

2(-1) = -2

-2 = -2

julsineya [31]3 years ago

6 0

Brackets first

2 * 6b is 12b
2 * 5 is 10
first part is 12b +10

The equation now is 12b + 10 is equal to 2 + 4b
take 2 from both sides
12b + 8 is equal to 4b
take 12b from both sides
8 is equal to -8b
divide by 8
1 is equal to -1b

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Hi pls help me thanksss<br>pls provide me with workings thankss :)

Vinil7 [7]

Step-by-step explanation:

(i)

Using cos(a - b) = cos a cos b + sin a sin b

$\frac{4}{5}$ = cos a cos b + $\frac{1}{5}$

cos a cos b = $\frac{4}{5} -\frac{1}{5} = \frac{3}{5}$

(ii)

Using cos(a + b) = cos a cos b - sin a sin b

cos(a + b) = $\frac{3}{5}-\frac{1}{5}=\frac{2}{5}$

(iii)

Using cot a = $\frac{cos a}{sin a}$

$=(\frac{cos A}{sin A} )(\frac{cos B}{sin B} )$

$=\frac{cos A cos B}{sin A sin B}$

$=\frac{\frac{3}{5} }{\frac{1}{5} }$ $=\frac{3}{5}$ × $5$

= 3

please give me a brainliest answer

8 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Write an equation in point slope form that passes through the point (1,8) & slope of -3 show all work

densk [106]

y-y1 = m(x-x1)

where: x1 = 1 y1 = 8 (given point (1,8) and m = -3

y - (8) = -3 (x - (1))

y - 8 = -3 (x-1)

y- 8 = -3x +1

6 0

3 years ago

Nvm bro i got the answer

KIM [24]

Ok and have a nice day

6 0

2 years ago

How does place value help me divide

Anna71 [15]

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3 years ago

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