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3 years ago

Brainliest for whoever gets this right Please help ASAP!!!

Mathematics

1 answer:

faust18 [17]3 years ago

4 0

Answer :

x = 2

y = -1

I hope it helps

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Give the sine, cosine, and tangent ratios for<br> both angle A and Angle B as a reduced fraction.

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Answer:

sin A= 43.60 or 43 3/5 cos A= 43.60 or 43 3/5 tan A= 43 3/5

sin B= 46. 40 or 46 2/5 cos B= 46. 40 or 46 2/5 tanB= 46.40 or 46 2/5

Step-by-step explanation:

Hope this Helped

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The answer to the triangle the hexagon and the Diamond

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1+3+3=7
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3 years ago

prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact

Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that limit property of an integral

$\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$

$\int_{a}^{b} fdt=f(b)-f(a)$ ....(Equation I)

Using above property of integral then we get

$\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$ ......(Equation II)

Substitute equation I and equation II are equal

Then we get

$\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}$

$\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)$

$\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)$

Therefore, $\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt$ .

Hence, the value of function does not change after changing the value of function at c.

6 0

3 years ago