Ask question

Ask question

All categories

4 years ago

6

Help!! Math question

2 answers:

aksik [14]4 years ago

5 0

18 - 7x = -20.52.5 = 7xx = 5/14 (c)

Zielflug [23.3K]4 years ago

3 0

18-7x=-20.5

-18 from both sides

Dive by -7 to both sides

D. 5 1/2

You might be interested in

A toy train is on a circular track. Luis lets the train run 60 degree around the track. Then he lets it run another 120 degree a

NeX [460]

The train would have traveled 40 degrees around the track.

7 0

3 years ago

Read 2 more answers

What is the average rate of change for this function for the interval from x = 2

andrew-mc [135]

B because (16-4)/2=6

7 0

4 years ago

Six times the quantity x -2 is 24. Find the number x

slamgirl [31]

The answer is 3.7 when rounded

8 0

3 years ago

Find the value(s) of the variable(s). Write your answer(s) in simplest radical form.

Romashka [77]

Answer:

Step-by-step explanation:

12). By applying Pythagoras theorem in the given right triangle,

Hypotenuse² = (Leg 1)² + (Leg 2)²

x² = 7² + 7²

x² = 2(7)²

x = $\sqrt{2(7)^{2} }$

x = $7\sqrt{2}$

13). By sine rule,

sin(60)° = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\frac{\sqrt{3}}{2}=\frac{9\sqrt{3} }{y}$

y = $\frac{18\sqrt{3} }{\sqrt{3} }$

y = 18

By cosine rule,

cos(60)° = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\frac{1}{2}=\frac{x}{y}$

x = $\frac{y}{2}$

x = $\frac{18}{2}=9$

14). By applying Pythagoras theorem in the given right triangle,

12² = x² + x²

144 = 2x²

72 = x²

x = √72

x = 6√2

5 0

4 years ago

Hi. I need help with these questions (see image)<br>Please show workings.<br>

pantera1 [17]

Answer:

see explanation

Step-by-step explanation:

Using the chain rule

Given

y = f(g(x)), then

$\frac{dy}{dx}$ = f'(g(x)) × g'(x) ← chain rule

and the standard derivatives

$\frac{d}{dx}$ ( $log_{a}$ x ) = $\frac{1}{xlna}$ , $\frac{d}{dx}$ (lnx) = $\frac{1}{x}$

(a)

Given

y = $log_{a}$ $\sqrt{(1+x)}$

$\frac{dy}{dx}$ = $\frac{1}{lna\sqrt{(1+x)} }$ × $\frac{d}{dx}$ ( $(1+x)^{\frac{1}{2} }$

= $\frac{1}{lna\sqrt{(1+x)} }$ × $\frac{1}{2}$ $(1+x)^{-\frac{1}{2} }$ × $\frac{d}{dx}$ (1 + x)

= $\frac{1}{lna\sqrt{(1+x)} }$ × $\frac{1}{2\sqrt{(1+x)} }$ × 1

= $\frac{1}{2lna(1+x)}$

= $\frac{1}{(1+x)lna^2}$

(b)

Given

y = ln sinx

$\frac{dy}{dx}$ = $\frac{1}{sinx}$ × $\frac{d}{dx}$ (sinx)

= $\frac{1}{sinx}$ × cosx

= $\frac{cosx}{sinx}$

= cotx

5 0

3 years ago