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3 years ago

What is the x-coordinate of the point (5,-3)? What is the y-coordinate?

Mathematics

2 answers:

AlladinOne [14]3 years ago

7 0

(x,y)

x-coordinate = 5
y-coordinate = -3

marissa [1.9K]3 years ago

4 0

$(x;\ y)\Rightarrow(5;-3)\Rightarrow x=5\ and\ y=-3\\\\Look\ at\ the\ picture$

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EASY MATH!!! GIVING BRAINLIEST!!

musickatia [10]

Answer:

Step-by-step explanation:

So lets take our 50 dollars.

Next, take your increase of 5%. Remember, 5% is 0.05, and it is a increase of 5%, so we are multipling 50 by 1.05.

Before we start multiplying, lets recall that this occurs twice(2 years), but this doesnt mean we multiply 50 by 1.10. No, we just mutiply 50 by 1.05 2 times:

50*1.05

52.5

52.5*1.05

55.25

So your answer is 55 dollars(rounding down).

Hope this helps!

4 0

2 years ago

Please help quick I will give brainliest :(

Flura [38]

Answer:

r=4 d=8 a = 16π c = 8π

d=6 r=3 a=9π c=6π

a=36π r=6 d=12 c=12π

c=18π r=9 d=18 a=81π

Step-by-step explanation:

hope this helps (:

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

2 years ago

Write the standard form of the equation of the line that passes through (3,2) and is parallel to the line whose equation is y=2x

Agata [3.3K]

Your answer is y=2x-4

3 0

3 years ago

What is the correct answer

anzhelika [568]

A and D
should be the answers.

5 0

3 years ago