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DerKrebs [107]
3 years ago
12

What is the value of b in the triangle shown below?

Mathematics
1 answer:
Jobisdone [24]3 years ago
4 0
The answer to the question is b.4in
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Plz help will make brainlyist
siniylev [52]

Answer:

A. 6

B. 143

C. The words "less than", "equal to", and "no more than"

Step-by-step explanation:

For a, we would want to first make an equation to represent the problem. It appears you've already figured that out though, so lets solve it!

Subtract 56 on both sides.

-6m=-36

Now divide -36 by -6 to isolate m.

m=6

It will take her 6 minutes.

For B, we need to make another equation.

2.95b<450-28

First, we can subtract 28 from the total, because she has already spent $28.

Now our equation is 2.95b<422

Lastly, just divide both sides by 2.95 to isolate b.

Our solution is 143. So she can buy 143 batteries.

Lastly, for 13, we can say that the words "less than", "equal to" and "no more than" can indicate an inequality in a real word problem.

Hope this helps!

3 0
3 years ago
PLS help I will give brainliest to the one who will answer me first.<br><br> pic is not blurry
kiruha [24]

Answer:

mean for a = 60/10 = 6

mad of a = 2

mean for b = 80/10 = 8

mad of b = 2

Step-by-step explanation:

Mean absolute deviation (MAD) of a data set is the average distance between each data value and the mean. Take each number in the data set, subtract the mean, and take the absolute value. Then take the sum of the absolute values. Now compute the mean absolute deviation by dividing the sum above by the total number of values in the data set. The mean absolute deviation, MAD, is 2.

\frac {1}{n} \sum \limits_{i=1}^n |x_i-m(X)|

m(X) = average value of the data set

n = number of data values

x_i = data values in the set

mean = average.

5 0
2 years ago
PLZ HELP
Gnesinka [82]

Answer:

Step-by-step explanation:

The highest power of x here is 3, so this is a third degree polynomial.  Since the coefficient of the highest power term is +, we know that the graph approximates that of the parent function y = x^3, and that this graph begins in Quadrant III, enters Quadrant I and continues to move upward in Quadrant I.

6 0
3 years ago
Read 2 more answers
Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
2 years ago
Which is the best estimate of -14 1/9 (-2 9/10)?<br>-42<br>-28<br>28<br>42​
andrew-mc [135]
-42 is the best estimate
5 0
2 years ago
Read 2 more answers
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