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3 years ago

Only the king of math can solve this and will give them brainliest

Mathematics

1 answer:

geniusboy [140]3 years ago

7 0

Answer:

<h3 /><h3 /><h2>Let the base of the triangle = X</h2>

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On Monday Kennedy spent $2 on lunch. On Tuesday she spent $1 on a snack on Wednesday, her sister gave her $3. Graph the integers

Agata [3.3K]

3because the kennedy spent 2 dollars on lunch and on monday she spent 1 dollar and than her sister gave her 3

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4 years ago

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What does the degree of a polynomial expression tell you about its related polynomial function? Explain your thinking. Give an e

ludmilkaskok [199]

The graph of function like that may may never cross the x-axis,

so the function could have no real zeros.

If it does have zeros, they will come in pairs,

because "what goes up must come down" and vice versa.

Of course the function cannot have more zeros than its degree.

For example, a polynomial function of degree 6 could have 0, 2, 4, or 6 real zeros.

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4 years ago

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Morton and Lyndal invested together in a new venture. Lyndal invested $500 more than Morton. If they invested $5000 together, ho

Kaylis [27]

Lyndal invested $3000 and Morton invested $2000

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3 years ago

Use integration by parts to find the integrals in Exercise.<br> ∫^3_0 3-x/3e^x dx.

Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

$du=1dx$

$du=dx$

$v'=\frac{1}{e^x}$

$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

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3 years ago

What is the answer (4×-12)+(-x+7)

Bogdan [553]

Answer:

x-41

Step-by-step explanation:

use the foil method

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3 years ago

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