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2 years ago

The point-slope equation of a line is y- Yo= m(x-xo), where m is the

Mathematics

1 answer:

Ainat [17]2 years ago

6 0

Answer:

Step-by-step explanation:

This equation can be reduced to：

y=m*x-m*x0+y0

so m is slope

Lalalalala end

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(9x2 + 10x + 4) − (9x2 + 5x − 1)

ch4aika [34]

For this case we must simplify the following expression:

$(9x ^ 2 + 10x + 4) - (9x ^ 2 + 5x-1)$

We eliminate the parentheses taking into account that:

$- * + = -\\- * - = +\\9x ^ 2 + 10x + 4-9x ^ 2-5x + 1 =$

We add similar terms taking into account that:

Equal signs are added and the same sign is placed.

Different signs are subtracted and the major sign is placed.

$9x ^ 2-9x ^ 2 + 10x-5x + 4 + 1 =\\5x + 5$

Answer:

$5x + 5$

3 0

2 years ago

Tiffany wants to buy a sweater with a selling price of $187. Today, the sweater is on sale for 31% off. What is the markdown? Wh

Akimi4 [234]

The sale price would be $129.03

7 0

2 years ago

Read 2 more answers

Solve for aaa: a-\left(-3.75\right)=6.11a−(−3.75)=6.11a, minus, left parenthesis, minus, 3, point, 75, right parenthesis, equals

Ganezh [65]

Answer:

Step-by-step explanation:

To isolate a, we add -3.75 to both sides.

a-(-3.75)=6.11

a-(-3.75)+(-3.75)=6.11+(-3.75)

a=6.11+(-3.75) Simplifying, we get:

a=2.36

(I got this from khan)

4 0

3 years ago

Please help with the questions in the image

iren2701 [21]

Step-by-step explanation:

This seems to be calculus 1.

Question a

We have $-2x^3 + 5x^2 + 9$

m = slope = derivative

Find the derivative / slope of $-2x^3 + 5x^2 + 9$

We do this by differentiating the polynomials. There are a few methods to do this but I am going to use the power rule, which we multiply the constant by the exponent on the variable and subtract one from the exponent.

$\frac{d}{dx}(-2x^3 + 5x^2 + 9)$

$-6x^2 + 10x$

$m = -6x^2 + 10x$ when x = a

Now that we have this information, we can answer question b

Question b

The tangent line for Point (1, 12)

First find the slope by using our derivative.

$m = -6x^2 + 10x$

$m = -6(1)^2 + 10(1)$

$m = -6 + 10x$

$m = 4$

Now that we have our slope, use point slope form to find our tangent line

$y - 12 = 4(x - 1)$

$y(x) = 4x + 8$

Now lets do the same for the Point (2, 13)

Find the slope at the point.

$m = -6x^2 + 10x$

$m = -6(2)^2 + 10(2)$

$m = -24 + 10$

$m = -4$

Now find the tangent line using point slope form of a line.

$y(x) - 13 = -4(x - 2)$

$y(x) = -4x + 21$

Now graph the lines, which I have done and you can see by viewing the image I have attached.

6 0

3 years ago

What is the slope of the line that passes through the pairs of points (-6,8) (2'3)

Alecsey [184]

Im pretty sure the answer is D, because if i used the right formula,
$\frac{y2 - y1}{x2 - x1}$
you should get this in the formula
$\frac{8 - 3}{2 - ( - 6)}$
$\frac{5}{8}$
a 2 subtracting a negative 6 makes it a positive, meaning your just adding 6 plus2, to get 8

6 0

3 years ago