In triangle ABC, AC is a median. The value of AC is: A. 34 B. 6 C. 28

(-9) + (-4) =

Gwar [14]

$\\ \bull\sf\longmapsto -9+(-4)=-9-4=-13$

$\\ \bull\sf\longmapsto +3-(-5)=3+5=8$

$\\ \bull\sf\longmapsto 2\times 2=4$

$\\ \bull\sf\longmapsto 36+4=40$

$\\ \bull\sf\longmapsto 5\times 7=35$

$\\ \bull\sf\longmapsto 8(9)=72$

$\\ \bull\sf\longmapsto -4+(-2)=-4-2=-6$

$\\ \bull\sf\longmapsto 3-(-6)=3+6=9$

$\\ \bull\sf\longmapsto -6(-3)=18$

$\\ \bull\sf\longmapsto -4+(-4)=-4-4=-8$

$\\ \bull\sf\longmapsto 5(7)=35$

$\\ \bull\sf\longmapsto 4-(-5)=4+5=9$

$\\ \bull\sf\longmapsto 4-3=1$

$\\ \bull\sf\longmapsto 10+5=15$

$\\ \bull\sf\longmapsto 3(8)=24$

7 0

3 years ago

What is the answer to this problem? 5[8-(9x-7)]+6x=0

julia-pushkina [17]

5[8-(9x-7)]+6x=0
5[8- 9x + 7]+6x=0
40 - 45x + 35 + 6x = 0
-39x = -75
x = 1 36/39
x = 1 12/13

3 0

4 years ago

77. the volume of a cube is increasing at a rate of <img src="https://tex.z-dn.net/?f=10%20%5Cmathrm%7B~cm%7D%5E%7B3%7D%20%2F%20

Colt1911 [192]

Answer:

$\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$

Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$