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2 years ago

What is the area of a rectangle with vertices at (7,3), (12,3), (12, 11), and (7,

Mathematics

2 answers:

NISA [10]2 years ago

6 0

Answer:

Step-by-step explanation:

Setler [38]2 years ago

6 0

Answer: 40 units

Step-by-step explanation:

(7,3) (12,3) on the same y coordinate, 5 squares apart

(12,3) (12,11) on the same x coordinate, 8 squares apart

8 × 5 = 40

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Solve the equation. StartFraction dx Over dt EndFraction equals StartFraction 1 Over x e Superscript t plus 7 x EndFraction An i

MAVERICK [17]

Answer:

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

Step-by-step explanation:

We are given that

$\frac{dx}{dt}=\frac{1}{xe^{t+7x}}$

We have to find the implicit function

Using separation variable method

$\frac{dx}{dt}=\frac{1}{xe^t\cdot e^{7x}}$

By using property $x^a\cdot x^y=x^{a+y}$

$xe^{7x}dx=e^{-t}dt$

By using property $\frac{1}{x^a}=x^{-a}$

Taking integration on both sides

$\int xe^{7x}dx=\int e^{-t}dt$

Parts integration method

$\int u\cdot v dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

By parts integration method

$x\int e^{7x}dx-\int (\frac{dx}{dx}\int e^{7x}dx)dx=-e^{-t}+C$

Using formula $\int e^{ax} dx=\frac{e^{ax}}{a}+C$

$\frac{xe^{7x}}{7}-\frac{1}{7}\int e^{7x}dx=-e^{-t}+C$

$\frac{xe^{7x}}{7}-\frac{1}{49}e^{7x}+e^{-t}=C$

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

We are given that

$F(x,t)=C$

$F(x,t)=\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

8 0

2 years ago

Please help me i really need help please

maw [93]

Answer:

39°

Step-by-step explanation:

IM assuming xyz is a triangle

180 - 113 - 28 = 67 - 28 = 39

3 0

2 years ago

Order these numbers from LEAST to GREATEST. 5/8 3/4 ,7/9

Paladinen [302]

The answer is C because 3/4 is the biggest of them all and the only one saying that is C.

3 0

2 years ago

WHERES THE BEEF???????????????????<br> hehehe sorry ab that im just HELLA bored over here ;)

ddd [48]

Me too tbhh school sucks omggg I have to put like an equation because If

5 0

2 years ago

I know the answer just need someone explain to me how to solve it

Vinvika [58]

The logarithm of a quotient can be written as a difference of logarithms:

$\log_3\left(\dfrac uv\right) = \log_3(u) - \log_3(v)$

You can also think of this as a combination of the product-to-sum and reciprocal/power properties of logarithms:

$\log_3\left(\dfrac uv\right) = \log_3\left(u \times \dfrac1v\right) = \log_3(u) + \log_3\left(\dfrac1v\right) = \log_3(u) - \log_3(v)$

To summarize,

$\log_b(mn) = \log_b(m)+\log_b(n)$

$\log_b\left(\dfrac mn\right) = \log_b(m) - \log_b(n)$

$\log_b\left(m^n\right) = n \log_b(m)$

3 0

1 year ago