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3 years ago

What are the constant terms in the expression? 8m−5n+2+m−14

Mathematics

2 answers:

Maslowich3 years ago

6 0

Answer: $2 , -14$

Step-by-step explanation:

We know that the constant terms are the specific terms in an algebraic expression that have only numbers or numerical value or without any variable.

For example : The constant term in $4x^2 -5m+16$ is 16.

Let's take one more example , the constant term in $3x-2y-4-7xy$ is -4.

Given expression is : $8m-5n+2+m-14$

Here, the terms are : $8m,\ -5n,\ 2,\ m,\ -14$

Terms having variable = $8m,\ -5n,\ m$

Terms without variable = $2 , -14$

⇒ Constant terms = $2 , -14$

Hence, the constant terms in the given expression = $2 , -14$

elena-s [515]3 years ago

5 0

The constant terms are: -14, and +2

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The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)

Lemur [1.5K]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

$\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt$

Do, integration by parts

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}$

Substitute the value in 1

Cardiac output: $F=\frac{3}{54.49}$

Cardiac output: $F=0.055 L\s$

Hence Cardiac output: $F=0.055 L\s$

4 0

2 years ago

Write the equation of a line that passes a given point and has the given slope. Answer in standard form. 1. Point (-3,5), m = -7

Likurg_2 [28]

Answer:

y -5 = -7(x +3)
y = 7

Step-by-step explanation:

The point-slope form of the equation can be used for this. It is generally written as ...

y -k = m(x -h) . . . . . line with slope m through point (h, k)

1. For m=-7, (h, k) = (-3, 5), the equation is ...

y -5 = -7(x +3)

2. For m=0, (h, k) = (4, 7), the equation is ...

y -7 = 0(x -4)

y = 7

_____

Comment on the form of the answer

Since a point and slope are given, "the equation" is appropriately written in point-slope form. For the second problem the simplified version would be ...

y -7 = 0

Sometimes, "the equation" means the answer is expected in slope-intercept form. In that case, a little rearranging is necessary for the first answer. It becomes ...

y = -7x -16

There are perhaps a dozen different forms that a linear equation can be written in. Considering those, "the equation" is somewhat ambiguous. Your curriculum materials may offer a clue as to the form intended.

5 0

3 years ago

2×+3 for x=3 I got 2 so I was seeing if I got it right

elena-s [515]

It would be 9 because 2 x 3 = 6 and then 6 + 3 = 9

7 0

3 years ago

Read 2 more answers

Find the slope of the line passing through the points (-7,-7) and (4,-7)

erastova [34]

Answer:

it it negative