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3 years ago

What equation contains the given point ( -3,5)

Mathematics

2 answers:

Alenkasestr [34]3 years ago

6 0

Answer:

y = x + 8

Step-by-step explanation:

attashe74 [19]3 years ago

5 0

Answer:

To determine if a point is on a line you can simply subsitute the x and y coordinates into the equation. Another way to solve the problem would be to graph the line and see if it falls on the line. Plugging in will give which is a true statement, so it is on the line.

Step-by-step explanation:

Off the inter net... sorry

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I paid $7.47 for 3 pounds of grapes. What is the cost of the grapes as a unit rate?

Vaselesa [24]

$7.47/3= $2.49
$2.49 per pound

6 0

3 years ago

Read 2 more answers

Which formula is best used to prove that a figure has congruent sides?

Mazyrski [523]

Answer:

Step-by-step explanation:

Congruent sides mean that they have "EQUAL" lengths.

So, to prove to sides are congruent means that their lengths are equal.

Slope gives the slope of a line. So this can't be it.

Area gives the area of figure. Nothing to do with congruency of 2 lines.

Midpoint formula gives the coordinates of the midpoint. Nothing here.

Distance formula gives the length of sides. So, this is the one we can use to prove that a figure has congruent sides. We can place the figure in coordinate system and find the length of 2 respective sides using "distance formula"

Correct answer is D

3 0

3 years ago

Given b(x)= (-3,7), state what the coordinates of this point will be of the following transformations are effected: 2b(x)

Dennis_Churaev [7]

The point translated up 7 and left 3. This will cause the point to go from (-3,7) to (-6,14)

3 0

3 years ago

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.