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3 years ago

What is the answer to -5x+5y=-10 and 3x-3y=5 using substitution

Mathematics

1 answer:

IceJOKER [234]3 years ago

7 0

Answer:

for the 1st one the answer is x= 3 y=1 and the second one is x= 5 and y = 3 and 1/3

Step-by-step explanation: so basically for the first one -5 times x which is 3 is -15 right? and then 5 times y which is 1 is 5 right so -15 plus 5 = -10

second one is x = 5 and y = 3 and 1/3 because 5*x which is 5 is = to 15 right? and then 3 times y which is equal to 3 and 1/3 is equal to 10 which when you do 15-10 in pretty sure it equals 5 correct me if im wrong

Hope this helps xD ;)

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In mathematics given that x=L2-L1/L1(O2-O1).

Assoli18 [71]

Answer:

1. L1 = $\frac{L2}{X(O2 -O1)}$

2. O2 = $\frac{L2 - L1(1 + XO1)}{XL1}$

Step-by-step explanation:

Given: x = $\frac{L2 - L1}{L1(O2 -O1)}$

1. To make L1 the subject of formula;

x = $\frac{L2 - L1}{L1(O2 -O1)}$

cross multiply to have;

L2 - L1 = xL1(O2 - O1)

collect like terms,

L2 = xL1(O2 - O1) + L1

factorize the right hand side;

L2 = L1[x(O2 - O1)]

L1 = $\frac{L2}{X(O2 -O1)}$

2. To make O2 the subject of formula;

x = $\frac{L2 - L1}{L1(O2 -O1)}$

cross multiply to have;

L2 - L1 = xL1(O2 - O1)

open the bracket to have;

L2 - L1 = xL1O2 - xL1O1

⇒ xL1O2 = L2 - L1 + xL1O1

O2 = $\frac{L2-L1 + XL1O1}{XL1}$

= $\frac{L2 - L1(1 + XO1)}{XL1}$

4 0

3 years ago

the radius of a sheridan balloon is increasing at a rate of 3 centimeters per minute. how fast is the volume changing when the r

san4es73 [151]

The volume of the balloon is $2352\pi cc/min$

Explanation:

The radius of the balloon is increasing at a rate of 3 cm/min.

To determine the volume of the balloon when the radius is 14 cm, we shall use the formula $V=\frac{4}{3} \pi r^3$

The rate of change of r with respect to time t is given by,

$\frac{d}{d t}(r)=3 \mathrm{cm} / \mathrm{minute}$

Now, we shall determine the $\frac{d}{d t}(V)$

$\begin{aligned}\frac{d}{d t}(V) &=\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right) \\&=\frac{4}{3} \pi\left(3 r^{2}\right)\frac{d}{d t}(r) \\&=4 \pi r^{2}\frac{d}{d t}(r)\end{aligned}$