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Marianna [84]
3 years ago
8

Pls answer i’ll give brainliest:)))

Mathematics
2 answers:
Tema [17]3 years ago
8 0
A - 1
B - 1
C - 3
Hope it helps :)
il63 [147K]3 years ago
7 0

Answer:

I give you 50 dollars if you brinly me you chose

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The ratio of the number of cookies in Jar A to the number of cookies in Jar B is 3:8. Half of the cookies in Jar A are transferr
motikmotik

The new ratio of the number of cookies in Jar A to the number of cookies in Jar B in the end is 5 : 19

<h3>How to find ratio</h3>

Ratio of jar A to jar B = 3 : 8

Jar B = 1/2 of A + B

= 1/2 × 3 + 8

= 3/2 + 8

= (3+16) /2

= 19/2

Jar A = 3 - 1/2

= (6-1) /2

= 5/2

New ratio of jar A to jar B = 5/2 : 19/2

= 5/2 ÷ 19/2

= 5/2 × 2/19

= 5/19

= 5 : 19

Learn more about ratio:

brainly.com/question/2784798

6 0
2 years ago
Create a function that has a derivative of f(x) = 2x + 1 at a given input value of your choice.
Taya2010 [7]
The opposite operation to taking a derivative is an integral. Integrate to find original function.

f(x) = x^2 + x + C

C is constant because we didn't have bounds on the integral. it lets you choose the input value let's say is just (1,0)

0 = 1 + 1 + C
- 2 = C

function at input value (1,0)
f(x) = x^2 + x - 2

now if you check by taking derivative is :
f' = 2x + 1
6 0
3 years ago
Find the area of ABC.
love history [14]

Answer:

Suppose, a triangle ABC, whose sides are a, b and c, respectively. Thus, the area of a triangle can be given by; Area= 30

Area = 1/2 x 15 x4 = 30 square yards

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please help due today!
wolverine [178]
Sorrysorrysorrysorrysorry
7 0
3 years ago
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Let z = ln(x 2 + y), x = ret . and y = ter . Use the Chain Rule to compute ∂z ∂r and ∂z ∂t at the point where (r, t) = (1, 2).\
Natali [406]

By the chain rule,

\dfrac{\partial z}{\partial u}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u}

where u\in\{r,t\}.

We have component partial derivatives

\dfrac{\partial z}{\partial x}=\dfrac{2x}{x^2+y}=\dfrac{2re^t}{r^2e^{2t}+te^r}

\dfrac{\partial z}{\partial y}=\dfrac1{x^2+y}=\dfrac1{r^2e^{2t}+te^r}

\dfrac{\partial x}{\partial r}=e^t

\dfrac{\partial x}{\partial t}=re^t

\dfrac{\partial y}{\partial r}=te^r

\dfrac{\partial y}{\partial t}=e^r

Putting the appropriate pieces together and setting (r,t)=(1,2), we get

\dfrac{\partial z}{\partial r}(1,2)=\dfrac{2e^3+2}{e^3+2}

\dfrac{\partial z}{\partial t}(1,2)=\dfrac{2e^3+1}{e^3+2}

3 0
3 years ago
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