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3 years ago

Solve (x + 5)2 = 48.

Mathematics

1 answer:

natita [175]3 years ago

7 0

Answer:

x = -5 ±4sqrt(3)

Step-by-step explanation:

(x + 5)^2 = 48

Take the square root of each side

sqrt((x + 5)^2) = ±sqrt(48)

x+5 = ±sqrt(16*3)

x+5 = ±4sqrt(3)

Subtract 5 from each side

x = -5 ±4sqrt(3)

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Two mechanics worked on a car. the first mechanic worked for 5 hours, and the second mechanic worked for 15 hours. together they

Ksenya-84 [330]

Alright, lets get started.

Suppose first mechanic charges= x

So, second mechanic charges will be = $(125 - x)$ (because total of their rates was 125)

first mechanic worked for 5 hours, and the second mechanic worked for 15 hours and together they charged a total of $1225

Means

$5 x + 15 (125-x) = 1225$

$5 x + 1875 - 15 x = 1225$

$1875 - 10 x = 1225$

Adding 10 x in both sides

$1875 - 10 x + 10 x = 1225 + 10 x \\1875 = 1225 + 10 x$

Subtracting 1225 from both sides

$1875 - 1225 = 1225 + 10 x - 1225\\650 = 10 x$

Dividing 10 in both sides

$\frac{650}{10} = \frac{10x}{10}$

x = 65

For second mecahnics rate = 125 - 65 = 60

Hence The mechanic who worked 5 hours charged $ 65 per hour, the one who worked 15 hours charged $ 60 per hour: Answer

Hope it will help :)

3 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

2 years ago

I need an answer asap

xenn [34]

25 ASAP miles per hour

8 0

3 years ago

Read 2 more answers

Please help I do not understand

pogonyaev

Answer:

-3/7

Step-by-step explanation:

Hi,

First of all, we know that the slope is not undefine because both the x-coordinates are not the same.

To find the slope, do change in y/change in x. Here, it would be ((2/7)-(5/7))/(2-1) or (-3/7)/1. Hence, the slope is -3/7.

8 0

2 years ago

Help

shepuryov [24]

Answer:

Step-by-step explanation:

g(x) = x-6

- g(x) = -x+6

g(x) + -g(x) = x-6 + -x+6

= x-x -6+6

4 0

3 years ago