All categories

3 years ago

Solve the following system algebraically:

Mathematics

2 answers:

kati45 [8]3 years ago

4 0

Guy above is right I’m pretty sure

Basile [38]3 years ago

3 0

Answer: A (8,4)

Step-by-step explanation:

X=8

2x+y=20

2(8)+y=20

16 +y = 20

Y= 4

(8,4)

You might be interested in

Given cos(angle) = -5/13 where the terminal arm of angle lies in quadrant 2, evaluate each trigonometric expression

Mrrafil [7]

Answer the answer is 76.6

Step-by-step explanation:

3 0

3 years ago

Can h help with this?

Setler [38]

The Answer would be p(m)=25+10 because she started with a base pay of 10 dollars and the next lawn she charge 35 meaning it makes the rate of change 25

6 0

3 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

3 years ago

Does anyone have answers for the Unit 6-Lesson 9: Rational Expressions and Functions Unit Test? I really need them, and will giv

GalinKa [24]

Answer:

on edge?

Step-by-step explanation:

5 0

3 years ago

How do you rationalise this surd?

Ksivusya [100]

$\frac{1 + \sqrt{3} }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{ \sqrt{2} + \sqrt{6} }{2}$
here you go

5 0

3 years ago

Read 2 more answers