All categories

3 years ago

3. Solve the following equation. * (r - 10) = -4

Mathematics

2 answers:

Kobotan [32]3 years ago

7 0

Answer:

R=6

Step-by-step explanation:

Du.mb_a_ss

Gala2k [10]3 years ago

7 0

Add 10 to both sides

so -4+10

=6

r=6

hope this helps

You might be interested in

Use a graph in a (-2π, 2π, π/2) by (-3, 3, 1) viewing rectangle to complete the identity.

yaroslaw [1]

First, notice that:

$2\tan (\frac{x}{2})=2\cdot(\pm\sqrt[]{\frac{1-cosx}{1+\cos x})}$

And in the denominator we have:

$1+\tan ^2(\frac{x}{2})=1+\frac{1-\cos x}{1+\cos x}=\frac{1+cosx+1-\cos x}{1+cosx}=\frac{2}{1+\cos x}$

then, we have on the original expression:

$\begin{gathered} \frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}=\frac{2\cdot\pm\sqrt[]{\frac{1-\cos x}{1+cosx}}}{\frac{2}{1+\cos x}}=\frac{2\cdot(\pm\sqrt[]{1-cosx})\cdot(1+\cos x)}{2\cdot(\sqrt[]{1+cosx})} \\ =(\sqrt[]{1-\cos x})\cdot(\sqrt[]{1+\cos x})=\sqrt[]{(1-\cos x)(1+\cos x)} \\ =\sqrt[]{1-\cos^2x}=\sqrt[]{\sin^2x}=\sin x \end{gathered}$

therefore, the identity equals to sinx

8 0

1 year ago

Which of the following is a true statement about skew lines?

otez555 [7]

A.) would be your answer

8 0

3 years ago

Read 2 more answers

Of the employees who work at stalling printing 90% attened the safety procedures meeting. If 63 employees attended the meeting h

My name is Ann [436]

90% of what is 63....

0.90x = 63
x = 63 / 0.90
x = 70....so 70 ppl work at stalling printing

6 0

3 years ago

HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH

8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

$4+\sqrt{6x}$

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

$4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}$

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

$\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}$

Thank you

8 0

3 years ago

Question. 1 :

vovikov84 [41]

<h3>Answer to Question 1:</h3>

AB= 24cm

BC = 7cm

<B = 90°

AC = ?

<h3>Using Pythagoras theorem :-</h3>

AC^2 = AB^2 + BC ^ 2

AC^2 = 24^2 + 7^2

AC^2 = 576 + 49

AC^2 = √625

AC = 25

<h3>Answer to Question 2 :-</h3>

sin A = 3/4

CosA = ?

TanA = ?

<h3>SinA = Opp. side/Hypotenuse</h3><h3> = 3/4</h3>

(Construct a triangle right angled at B with one side BC of 3cm and hypotenuse AC of 4cm.)

<h3>Using Pythagoras theorem :-</h3>

AC^2 = AB^2 + BC ^ 2

4² = AB² + 3²

16 = AB + 9

AB = √7cm

<h3>CosA = Adjacent side/Hypotenuse</h3>

= AB/AC

= √7/4

<h3>TanA= Opp. side/Adjacent side</h3>

=BC/AB

= 3/√7

4 0

2 years ago