All categories

3 years ago

Which ordered pair is a solution of the equation? 7x-2y=-5

2 answers:

8 0

answer X= 5 over 7 and 2 over 7y step by step explaination add 2y on both sides u cancelled -2y and positive 2y u bring down 7X u get 7X=-5+3 then u divide and u get that answer yw have a nice day mark me brainliest

laiz [17]3 years ago

5 0

Answer:

2y=-x7

Love mORgan

You might be interested in

2. Help ASAP!!!!!!!!!!!!!!!!

ch4aika [34]

ANSWER

$2 \times 2$

EXPLANATION

The dimensions of a matrix is of the form:

$Row \times Column$

The dimensions of the first matrix is

$2 \times 3$

The dimensions of the second matrix is

$3 \times 2$

Since the inner products are the same,the outer products give the dimensions of the product of the two matrices.

$(2 \times 3)(3 \times 2) = 2 \times 2$

Therefore the dimensions of the product is a 2×2 matrix.

8 0

3 years ago

Read 2 more answers

Which function is graphed?

Alexandra [31]

Hey there! :)

Answer:

C. {x² + 4, x < 2

{-x + 4 x ≥ 2

Step-by-step explanation:

This is a piecewise function, where the two equations are different. They are:

y = x²+ 4

y = -x + 4

The function x² + 4 is graphed where x < 2. (< is used because the circle is open)

The function -x + 4 is graphed where x ≥ 2. (≥ is used since the endpoint is closed)

Therefore, the correct answer is:

C. x² + 4, x < 2

-x + 4 x ≥ 2

5 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

2 years ago

How much are these coins worth?

deff fn [24]

Answer:

Step-by-step explanation:

1.65

Things how much there worth

Hope it helps

3 0

2 years ago

Marcus has $45 in his bank account. He deposits a check for $65. Two days later he

Nonamiya [84]

There is $90 left in his account

7 0

3 years ago

Read 2 more answers