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2 years ago

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The numbers in the diagram below indicate the lengths of the sides of a triangle. Bernadette drew the following triangle and cla

ims it is a right triangle.
State whether or not Bernadette is correct?

1 answer:

xz_007 [3.2K]2 years ago

8 0

Answer:

It's not a right triangle, there isn't a right angle anywhere there.

Also, it's not applicable via pythagorean theorem

Step-by-step explanation:

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Factorise completely 15a^3 - 5ab<br><br>pwease help me

LenKa [72]

The first step to factorize is to find the common terms. In this 5a is common throughout. By taking it out we get the expression:

5a (3 a^2 -b)

Hope this helps!

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2 years ago

An ecologist took samples of several different tundra plants.

Annette [7]

The answer of this question is 6/19. I hope this helps

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3 years ago

Read 2 more answers

PLeAsE i NeEd HeLp!!!!

olya-2409 [2.1K]

Answer:

<em>DF = 10 units</em>

$m\angle DFG = 28^\circ$

<em>EG = 5.04</em>

Step-by-step explanation:

<u>Properties of Rhombus es</u>

All Sides Of The Rhombus Are Equal.
The Opposite Sides Of A Rhombus Are Parallel.
Opposite Angles Of A Rhombus Are Equal.
In A Rhombus, Diagonals Bisect Each Other At Right Angles.
Diagonals Bisect The Angles Of A Rhombus.

The image contains a rhombus with the following data (assume the center as point O):

DO = 5 units

GF = 5.6 units

$m\angle FEO = 62^\circ$

4. Calculate DF

Applying property 4, diagonals bisect each other, thus the length of DF is double the length of DO, i.e. DF=2*5 = 10:

DF = 10 units

5. Calculate $m\angle DFG$

Applying property 4 in triangle EFO, the center angle is 90°, thus angle EFO has a measure of 90°-62°=28°.

Applying property 5, this angle is half of the measure of angle EFG and angle DFG has the same measure of 28°.

$m\angle DFG = 28^\circ$

6. FG is the hypotenuse of triangle OFG, thus:

$OG^2=FG^2-OF^2$

$OG^2=5.6^2-5^2$

$OG^2=6.36$

$OG=\sqrt{6.36}=2.52$

EG is double OG: OG=2*2.52=5.04

EG = 5.04

5 0

2 years ago

If the length of the minute hand of the clock in London commonly known as “Big Ben” is 11.25’ from the center of the clock to th

Elenna [48]

The angle swept out is equal to 1/12 of a full rotation is π/6 radians.

r^2 * theta = π/6 * 11.25^2 = 66.268"

3 0

3 years ago

addison and kelsey are running on a path modeled by x^2+y^2-10x-18y-378=0, where the distance is in meters. what is the maximum

bulgar [2K]

The maximum distance is the <u>diameter of the circle</u>, which is of 44 units.

The equation of a circle of <u>radius r and center</u> $(x_0,y_0)$ is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

The diameter is <u>twice the radius</u>, and is the <u>maximum distance</u> between two points inside a circle.

In this problem, the circular path is modeled by:

$x^2 + y^2 - 10x - 18y - 378 = 0$

We complete the squares to place it in the standard format, thus:

$x^2 - 10x + y^2 - 18y = 378$

$(x - 5)^2 + (y - 9)^2 = 378 + 25 + 81$

$(x - 5)^2 + (y - 9)^2 = 484$

Thus, the radius is:

$r^2 = 484 \rightarrow r = \sqrt{484} = 22$

Then, the diameter is:

$d = 2r = 2(22) = 44$

The maximum distance is the <u>diameter of the circle</u>, which is of 44 units.

A similar problem is given at brainly.com/question/24992361

7 0

2 years ago