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3 years ago

Help please! Will award brainliest!

Mathematics

1 answer:

emmasim [6.3K]3 years ago

5 0

C. (2x - 5)

Explanation:

Given:

The factors of $4x^2 - 25$ = ?

The binomial factor can be simplified and written as

$4x^2 - 25 = (2x + 5) (2x - 5)$

When these two factors are multiplied then they give 4x² - 25

Thus, (2x + 5) and (2x - 5) are the two factors of 4x² - 25

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Solution n/8+1/2 = -3 1/2

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Answer:

n= -32

Step-by-step explanation:

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2 years ago

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Look at picture for problem

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Add 2 3/4 and 2 1/2 and you’ll get ur answer. area means total and when u see area, it means add

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2 years ago

4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia

Ad libitum [116K]

The standard form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its standard form from the coordinates of the foci and a general equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the general equation of the circle into its standard form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (semi-major distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the standard form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

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1 year ago

An airline allows each passenger 33 kilograms of luggage. Carl has to lower the weight of his suitcase by 25% to stay within the

damaskus [11]

1- 0.25= 0.75 (multiplier)

33÷0.75=44

The suitcase originally weighed 44kg

Hope this helps!

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2 years ago

Read 2 more answers

Construct a quadrilateral MINT such that MI= 5cm, IN = 6cm, NT = 7cm, TM=3cm and MN =9cm

Tems11 [23]

From the given dimensions, of MI, IN, NT, TM, and MN, the quadrilateral

MINT can be drawn as shown in the attached image.

<h3>What are the steps for the construction of MINT?</h3>

The given dimensions of the quadrilateral MINT are;

MI = 5 cm

IN = 6 cm

NT = 7 cm

TM = 3 cm

MN = 9 cm

The side MN is a diagonal of MINT, therefore;

ΔMIN, and ΔMTN are triangles with a common base = MN

The steps to construct MINT are therefore;

Step 1; Draw the line MN = 9 cm.

Step 2; Place the compass at point M and with a radius MI = 5 cm, draw an arc on one side of MN.

Step 3; Place the compass at N and with radius IN = 6 cm, draw an arc to intersect the arc dawn in step 1 above.

Step 4; Place an arc at point M and with radius TM = 3 cm draw an arc on the other side of MN.

Step 5; Place the compass at point N and with radius NT = 7 cm, draw an arc to intersect the arc drawn in step 3.

Step 6; Join the point of intersection of the arcs to points M and N to complete the quadrilateral MINT.

Please find attached the drawing (showing the construction arcs) of the

quadrilateral MINT created with MS Word.

Learn more about types of geometric construction here:

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1 year ago