Solve for qqq. 3\left(q+\dfrac43\right) = 23(q+ 3 4 )=2pls answer this

Mathematics

1 answer:

enyata [817]3 years ago

6 0

Answer:

Step-by-step explanation:

Given the expression $3\left(q+\dfrac43\right) = 23$ , we are to find the value of q;

$3\left(q+\dfrac43\right) = 23\\on\ expansion\\\\3q + 4/3(3) = 23\\\\3q+4 = 23\\\\subtract \ 4\ from \ both\ sides \ of \ the \ equation\\\\3q+4-4 = 23-4\\\\3q = 19\\\\Diviide \both\ sides \ by \ 3\\\\3q/3 = 19/3\\\\q = 19/3$

Hence the value of q is 19/3

You might be interested in

Please help.<br>find the missing side or angle and each problem . <br>

Jlenok [28]

1. 27
2. 55
3. 20
4. 21

4 0

3 years ago

What is the remainder of (4x2 + 7x-1)= (4 + x)?<br>A. -9x – 1<br>B.23x – 1<br>C.35<br>D.-37

bearhunter [10]

Answer: I don't know what you meant by remainder but i hope this helps :)

$x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}\\$

Step-by-step explanation:

$\left(4x^2+7x-1\right)=\left(4+x\right)\\\mathrm{Refine}\\4x^2+7x-1=4+x\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\4x^2+7x-1-x=4+x-x\\Simplify\\4x^2+6x-1=4\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\4x^2+6x-1-4=4-4\\\mathrm{Simplify}\\4x^2+6x-5=0\\\mathrm{For\:}\quad a=4,\:b=6,\:c=-5:\\\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}$

$\frac{-6+\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\=\frac{-6+\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{8}\\\\Let\: simplify\: ; -6+2\sqrt{29}\\=-2\times \:3+2\sqrt{29}\\=2\left(-3+\sqrt{29}\right)\\=\frac{2\left(-3+\sqrt{29}\right)}{8}\\=\frac{-3+\sqrt{29}}{4}\\$

$\frac{-6-\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\\\=\frac{-6-\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\\\=\frac{-6-\sqrt{116}}{2\times \:4}\\\\=\frac{-6-2\sqrt{29}}{8}\\\\=-\frac{2\left(3+\sqrt{29}\right)}{8}\\\\=-\frac{3+\sqrt{29}}{4}\\\\\\x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}$