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2 years ago

What is the absolute value of -25?

Mathematics

2 answers:

zvonat [6]2 years ago

7 0

Answer:

Step-by-step explanation:

Absolute value: |-25|

Get rid of the negative.

You will get 25.

NISA [10]2 years ago

6 0

Answer:

Step-by-step explanation:

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The graph of f(x) = x2 is translated to form g(x) = (x – 5)2 + 1. On a coordinate plane, a parabola, labeled f of x, opens up. I

Firdavs [7]

The quadratic function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).

<h3>How to analyze quadratic equations</h3>

In this question we have a graph of a quadratic equation translated to another place of a Cartesian plane, whose form coincides with the vertex form of the equation of the parabola, whose form is:

g(x) = C · (x - h)² - k (1)

Where:

(h, k) - Vertex coordinates
C - Vertex constant

By direct comparison we notice that (h, k) = (5, 1) and C = 1. Now we proceed to check if the points (x, y) = (2, 10) and (x, y) = (8, 10) belong to the parabola.

x = 2

g(2) = (2 - 5)² + 1

g(2) = 10

x = 8

g(8) = (8 - 5)² + 1

g(8) = 10

The quadratic function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).

To learn more on parabolae: brainly.com/question/21685473

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2 years ago

An investment had return of 13% if inflation was 4.1% what was your real return ?

e-lub [12.9K]

Answer: 8.9%

Step-by-step explanation: 13 - 4.1

8 0

2 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago

find two consecutive odd integers such as 65 more than the lesser is four times the greater. The lesser consecutive odd integer

Alex_Xolod [135]

Answer: Lesser = 19, Greater = 21

See picture reply.

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3 years ago

Which is the interval notation represent the domain?

Kazeer [188]

Answer:

The first choice, $(-\infty, 7) \cup (7, \infty)$ .

Step-by-step explanation:

The dashed vertical line is a vertical asymptote. The x-coordinate of all points on this vertical asymptote are equal to 7. In other words, when the x-value of a point on the graph approaches $7$ , its y-value approaches infinity (or negative infinity.)

Either way, the graph is not defined for $x = 7$ . The point $7$ should thus be excluded from the domain of the graph.

The graph is apparently defined for all other x-values. The domain of the function should thus be all real numbers with the exception of $x = 7$ . Here's how to write that in interval notation:

The set of all real numbers can be expressed as the interval $(-\infty, \infty)$ . Note that infinity $\infty$ (or $(- \infty)$ itself isn't a real number. The round brackets indicate that both endpoints are excluded from the from the interval.
The set of all real numbers less than (not equal to) $7$ is $(-\infty,7 )$ (both ends are excluded.) The set of all real numbers greater than (not equal to) $7$ is $(7, \infty)$ .
The set of all real numbers that is not equal to $7$ is the union of all real numbers less than $7$ and all real numbers greater than $7$ . The union operator $\cup$ connects two intervals.

In other words, the domain is $(-\infty, 7) \cup (7, \infty)$ .

5 0

3 years ago