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ASHA 777 [7]
3 years ago
11

What is the constant of proportionality of 8:5?

Mathematics
1 answer:
Viktor [21]3 years ago
7 0
I am wondering is that 8/5??
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lubasha [3.4K]
-6 is the answer since 6*-1 is -6
4 0
3 years ago
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
(28 POINTS)
cestrela7 [59]

Answer:

(a) -\sqrt{10}y

(b) -\sqrt{3}y

Step-by-step explanation:

(a) \sqrt{10y^{2}} can also be written as \sqrt{10} x \sqrt{y^{2}}

the \sqrt{y^{2}} = ± y

Therefore :

\sqrt{10y^{2}} = ±\sqrt{10}y

since it was given that y < 0 , then the factor will be -\sqrt{10}y

(b) \sqrt{3y^{2}} can also be written as : \sqrt{3} x \sqrt{y^{2}}

Therefore , one of the factor is -\sqrt{3}y

3 0
3 years ago
What is the value of y in the triangle?
hoa [83]
Did you ever get the answer?

5 0
3 years ago
Somthing ab a mark up price, idrk tbh
gregori [183]

Since they reduced the markup price, or higher price, by 10%, the new markup price is 20% over the manufacturer price.

3 0
2 years ago
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