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sleet_krkn [62]
3 years ago
5

Please help me with this

Mathematics
1 answer:
NemiM [27]3 years ago
8 0
Google it and see what you get
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1.5.3
SIZIF [17.4K]

Answer:

A. x^2 = (4x - 1)^2 + (6x + 7)^2

Step-by-step explanation:

The question seems to be incomplete. However, we'll make the following assumption.

Opposite = 6x + 7

Adjacent = 4x - 1

Hypothenuse = x

Pythagoras theorem states that:

Hypothenuse^2 = Opposite^2 + Adjacent^2

Substitute values for Hypothenuse, Opposite and Adjacent.

This gives

x^2 = (6x + 7)^2 + (4x - 1)^2

or

x^2 = (4x - 1)^2 + (6x + 7)^2

From the list of given options.

A answers the question

5 0
3 years ago
If the diameter of a circle with a circumference of 19.5 inches is reduced by three, what would be the circumference of the new
Elanso [62]

It is given that the circumference of the circle is 19.5 inches. Let the diameter is d inches .

And the formula of circumference is

C= \pi d

Substituting the value of C, we will get

19.5= \pi d
\\
d = \frac{19.5}{ \pi}

When the diameter increased by 3, then the circumference is

C = \pi ( \frac{19.5}{ \pi} +3})=Approx 29 inches

And the circumference , when diameter is increased by 3 is 29 inches .

3 0
3 years ago
Read 2 more answers
(6^2)^4 simplify expression
grin007 [14]

Answer:

1679616

Step-by-step explanation:

6 0
3 years ago
The length of a rectangle is 3 times longer than its width. If the area of the rectangle is 24 square inches, then what is the p
NikAS [45]

Answer:  The answer is 16√2 cm.


Step-by-step explanation:  Given that there is a rectangle with length 3 times than its width. We are to find the perimeter of the rectangle.

Let 'w' represents the width of the rectangle.

Then, its length will be 3w.

Also, area of the rectangle = 24 square inches.

Therefore,

3w\times w=24\\\\\Rightarrow 3w^2=24\\\\\Rightarrow w^2=8\\\\\Rightarrow w=2\sqrt 2.

So, width = 2√2 inches and length = 6√2 inches.

Thus, perimeter of the rectangle = 4√2 + 12√2 = 16√2 cm.

4 0
3 years ago
Read 2 more answers
Cora is using successive approximations to estimate a positive solution to
Advocard [28]
Um...... what......?
7 0
3 years ago
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