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gogolik [260]
3 years ago
8

Is this function linear or nonlinear?​

Mathematics
1 answer:
sasho [114]3 years ago
3 0

Answer:

It's linear

Step-by-step explanation:

This is a slope intercept form.

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One more - I am very confused right now.
stira [4]

In order to find the length of the base we have to find x.

Now we know that an isosceles triangle has two sides that are congruent and a base.

We know that the two sides are x +1 and -x + 7

We can figure out x by making these two equal each other.

x + 1 = -x + 7  *first step is to get you x's together* 

+x     = +x    *Now the x's cancel on the right side. and group them on the left*

2x + 1 = 7  * Now subtract 1 form both sides*

     - 1   = - 1   * Solve the right side.

2x  = 6  * divide both sides by 2 to isolate the x to be alone*

÷2  = ÷2  * solve the right side.

x =  3


Now we know x lets plug in that number for the Length of the base.

3(3) - 3

9 - 3 = 6

So your answer is B. 6


Good Luck! :)

8 0
3 years ago
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HELP me please I don't know what to do
Ostrovityanka [42]
Every month, she'd pay $494.83.
7 0
4 years ago
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An item is priced at $13.86. If the sales tax is 6%, what does the item cost including sales tax?
Mazyrski [523]

Answer:

$14.69 rounded to the nearest cent would be $14.70, so you muitiply 13.86 x 6 percent which gives you 0.8316, so you take that and add it back to 13.86


3 0
3 years ago
Ben, Shaneeka, and Phil live on the same street. Ben lives 6/11 mile north of Phil, and 1/3 mile north of Shaneeka. How far does
Norma-Jean [14]

Answer:

Shaneeka live from Phil 7/33 mile far.

Step-by-step explanation:

Given:

Ben lives 6/11 mile north of Phil, and 1/3 mile north of Shaneeka.

Now, to get how far does Shaneeka live from Phil.

So, by subtracting the distance from  6/11 mile north where Ben lives from Phil to  1/3 mile north where Ben lives from Shaneeka:

\frac{6}{11} -\frac{1}{3}

=\frac{6\times 3 -1\times 11}{33}

=\frac{18-11}{33}

=\frac{7}{33}

Therefore, Shaneeka live from Phil 7/33 mile far.

7 0
3 years ago
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(O) = 2.8 y(t) =( Preview
Karolina [17]

Answer:

y=\frac{4}{5}e^{t}-\frac{4}{5}e^{-\frac{2}{5}t}

Step-by-step explanation:

The given equation 5y'' + 3y' - 2y =0 can be written as

(5D^{2}+3D-2)y(t)=0

Solving for complementary function we have Roots of (5D^{2}+3D-2) as follows

(5D^{2}+5D-2D-2)

5D(D+1)-2(D+1)=0\\\\(5D-2)(D+1)=0\\\\\therefore D=-1\\D=+2/5

Thus the complementary function becomes

y=y=c_{1}e^{m_{1}t}+c_{2}e^{m_{2}t}

where

m_{1},m_{2} are calculated roots

thus solution becomes

y=c_{1}e^{-t}+c_{2}e^{\frac{2}{5}t}

Now to solve for the coefficients we use the given boundary conditions

y(0)=0\\\\\therefore c_{1}+c_{2}=0\\\\y'(0)=-c_{1}+\frac{2}{5}c_{2}=2.8\\\\\therefore c_{2}+\frac{2}{5}c_{2}=2.8\\\\c_{2}=2\\\\\therefore c_{1}=-2}

hence the solution becomes

y=-2e^-{t}+2e^{\frac{2}{5}t}

8 0
3 years ago
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