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2 years ago

9

Point c is located at (1, -2) on the coordinate plane. Point c is reflected over the y -axis to create point c'. What ordered pa

ir describes the location of c'?

1 answer:

castortr0y [4]2 years ago

5 0

Answer:

(-1, -2)

Step-by-step explanation:

If it is reflected in the Y-Axis the X-Axis will be opposite to its original value.

Mark me as brainliest if you want to.

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Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

Tell which number is greater.<br><br> 76%, 0.67

Mariana [72]

Answer:

76%

Step-by-step explanation:

first use the calculator you might find the answer on it even the calculator on your own phone

sorry i know you wont understand me

8 0

3 years ago

60 Points !!!!!

zheka24 [161]

Answer:

A yes ; it can be rotated 360° or less and match the original figure

8 0

2 years ago

What is the angle B measurement. Urgent! Brainliest!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

forsale [732]

Answer:

52°

Step-by-step explanation:

both the triangle and the line equal 180 so, 180-127=53 53+75=128 180-128=52

8 0

3 years ago

Solve for the roots in the equation below. x4 + 3x2 - 4 = 0

fgiga [73]

First, you need to rewrite the expression into binomial form, so you are working with two terms (as you world with a quadratic):
(x²)²-3(x²)-4=0
Now, you can place the x²s into brackets as the coefficient is now 1:
(x² )(x² )
Next, find out two numbers that add together to give you -3 and multiply to give -4 (these are the leftover integers after removing the x²s). These two numbers are -4 and 1.
Place the -4 and 1 into the brackets:
(x²-4)(x²+1)=0
Notice that the x²-4 is a difference of two squares, so can be further factorised into (x+2)(x-2)
This leaves you with a final factorisation of:
(x+2)(x-2)(x²+1)=0
Now we handle each bracket individually to obtain our four solutions for x:
x+2=0
x=-2
x-2=0
x=2
x²+1=0
x²=1
x=<span>±1</span>

8 0

3 years ago