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3 years ago

To convert 1.3 miles into inches which two ratios could you multiply by

1 answer:

4 0

To convert 1.3 miles into inches we proceed as follows;
1 mile=63360 inches.
Therefore the number of inches in one mile will be as follows:
(1.3*63360)
=82,368 inches
Therefore, we conclude that our answer is 82,368 inches.

You might be interested in

Rewrite using a single positive exponent. 3-6 -5

yarga [219]

Answer:

$1 \div {3}^{( 30)}$

Step-by-step explanation:

There are exponent rules I posted above that are being used here.

We have (3^-6)^-5

Since we have a power times a power, we multiply 6 and 5 to get -30. Look at the first example in the photo.

So, we're left with 3^-30.

If you look at the photo, there's a rule for negative exponents.

Take the inverse of the base number, so 3 in this case:

1 / 3

and then make the exponent positive after taking the inverse.

So, we should get 1 / 3^30

Hope this helps!

3 0

1 year ago

A cylinder has a circular base with a radius that measures 9 inches. Determine the diameter of the base of the cylinder. 3 inche

yarga [219]

It is 18 because the radius is half the diameter and 9 x 2 = 18.

4 0

3 years ago

Read 2 more answers

Need help ASAP Tysm! :))

Mrrafil [7]

Answer:

The answer is -86

Step-by-step explanation:

-7 squared is 49

49 × 2 = 98

12-98= -86

4 0

3 years ago

Prove that<br> cos^4∆ - sin^4∆ = 2cos^2∆ - 1

GarryVolchara [31]

$\cos^{4} x-\sin^{4} x\\\\=(\cos^{2} x+\sin^{2} x)(\cos^{2} x-\sin^{2} x)\\\\=\cos^{2} x-\sin^{2} x\\\\=\cos^{2} x-(1-\cos^{2} x)\\\\=2\cos^{2} x-1$

3 0

1 year ago

Which set of parametric equations over the interval 0 ≤ t ≤ 1 defines a line segment with initial point (–5, 3) and terminal poi

Ksivusya [100]

Given:

A line segment with initial point (–5, 3) and terminal point (1, –6).

To find:

The set of parametric equations over the interval 0 ≤ t ≤ 1 which defines the given line segment.

Solution:

Initial point is (–5, 3). So,

$x(0)=-5,y(0)=3$

Terminal point is (1, –6).

$x(1)=1,y(1)=-6$

Check which of the given set of parametric equations satisfy $x(0)=-5,y(0)=3,x(1)=1,y(1)=-6$ .

Put t=1 in each set of parametric equations.

In option A,

$y(1)=3-6(1)=3-6=-3\neq -6$

So, option A is incorrect.

In option B,

$y(1)=1-6(1)=1-6=-5\neq -6$

So, option B is incorrect.

In option C,

$y(1)=3-9(1)=3-9=-6$

$x(1)=-5+6(1)=-5+6=1$

Put t=0, in this set of parametric equations.

$x(0)=-5+6(0)=-5$

$y(0)=3-9(0)=3$

So, option C is correct.

In option D,

$y(1)=1-7(1)=1-7=-3\neq -6$

$x(1)=-5+8(1)=-5+8=3\neq 1$

So, option D is incorrect.

8 0

2 years ago