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r-ruslan [8.4K]

3 years ago

14

A number m plus six times n

1 answer:

ikadub [295]3 years ago

6 0

Answer:

Xm + 6 * n

Im pretty sure from how I read it this would be the equation.

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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B400%20%5Ctimes%20400%20-%2024%20%5Ctimes%20400%20-%20360%20%5Ctimes%20400%20-%202

max2010maxim [7]

Answer:

$\sqrt{400 \times 400 - 24 \times 400 - 360 \times 400 - 200} \\ \\ = \sqrt{160000 - 9600 - 144000 - 200} \\ \\ = \sqrt{160000 - 9600 - 144000 - 200} \\ \\ = \sqrt{160000 - 153,800} \\ \\ = \sqrt{6200} \\ \\ = 78.74 \\ \\$

7 0

3 years ago

Question is in picture.

Norma-Jean [14]

Answer:

<h2>B. 2x + y = 4</h2>

Step-by-step explanation:

Having the system of equations in its simplest form

$\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right$

If

$a=d,\ b=e,\ c=f$

then the system of equations has infinitely many solutions.

If

$a=d,\ b=e,\ c\neq f$

then the system of equations has no solution.

If

$a\neq d\ or\ b\neq e$

then the system of equations has one solution.

We have the equation

$y=-2x+4$

Convert to the standard form Ax + By = C<em>:</em>

<em /> $y=-2x+4$ <em> add 2x to both sides</em>

$y+2x=-2x+2x+4\\\\2x+y=4$

3 0

4 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

3 years ago

mr wicks designs houses.he uses grid paper to plan a new house design.the kitchen will have and are between 70 square feet and 8

exis [7]

Well we can't draw on brainly so no.

4 0

4 years ago

Three years ago the average price of a movie ticket was $8.75 and now it is $11.00. What was the annual multiplier and the perce

Andrei [34K]

Given:
Price of tickets:
3 years ago - 8.75
now - 11

annual multiplier:

b³ = 11/8.75
b = ∛11/8.75
b = 1.079

The annual multiplier is 1.079.

1.079 * 100% = 107.9%

107.9% - 100% = 7.9%

The percentage increase is 7.9%

8 0

4 years ago