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3 years ago

Choose the steps that show how to solve this problem using arithmetic

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

4 0

Answer:

1) find d or distance between numbers

2) equation: an=a1+(n-1)d

3) an is nth term

a1 is first number in sequence

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A football field is rectangular and measures 360 feet by 160 feet. A flying disk is randomly tossed onto the field. If it is equ

kaheart [24]

The dimensions if the field are: the length is 360 feet; the width is 160 feet.

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3 years ago

Gnoma [55]

Answer: Third option.

Step-by-step explanation:

These are some transformations for a function f(x):

If $f(x)-k$ , then the function is shifted down "k" units.

If $f(x-k)$ , then the function is shifted right "k" units.

Knowing this we can describe the transformation from the graph of the function $f(x) = x^2$ to the graph of the function $f(x) = (x - 3)^2-1$ . This is:

The function $f(x) = (x - 3)^2-1$ is the function $f(x) = x^2$ but shifted right 3 units and shifted down 1 units.

Therefore, the correct option is the third one.

7 0

4 years ago

Read 2 more answers

In a quadrilateral ABCD, angle A + angle D = 90°. Prove that AC^2 + BD^2 = BC^2 + AD^2

olga2289 [7]

Answer:

$AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}$

$AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}$

Hence prove.

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

Step-by-step explanation:

Given:

∠A + ∠D = 90°

We are prove to that

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

Solution:

See required figure in attached file.

We know sum of the all angles of a triangle is 180°.

So, in triangle AED.

∠A + ∠E + ∠D = 180°

∠E + (∠A + ∠D) = 180°

Now, we substitute ∠A + ∠D = 90° in above equation.

∠E + 90° = 180°

∠E = 180° - 90°

∠E = 90°

Using Pythagoras Theorem for triangle ADE and triangle BEC.

$AD^{2}=AE^{2}+ED^{2}$

$BC^{2}=BE^{2}+EC^{2}$

Now, we add both above equations.

$AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}$ --------(1)

Similarly, Using Pythagoras Theorem for triangle AEC and triangle BED.

$AC^{2}=AE^{2}+EC^{2}$

$BD^{2}=BE^{2}+ED^{2}$

Now, we add both above equations.

$AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}$ --------(2)

We get From equation 1 and equation 2.

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

Hence prove,

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

4 0

3 years ago

What is the long division for 315 divided by 15

AleksAgata [21]

Simply divide 315 by 15= 21<u />

hard to illustrate but first 31/15= 2 remainder 1
15/15 =1

answer =21
hope this helps

7 0

3 years ago

Read 2 more answers

Find the sum. Write the sum as a mixed number, so the fractional part is less than 1. Number 3. 2 2/3 + 3 2/3 number 5. 9 3/6 +

pychu [463]

Answer:

See explanations below

Step-by-step explanation:

1)2 2/3 + 3 2/3

Convert to improper fraction

= 8/3 + 11/3

= (8+11)/3

= 19/3

= 6 1/3

hence the sum as mixed fraction is 6 1/3

2) 9 3/6 + 2 2/6

= 57/6 + 14/6

= (57+14)/6

= 71/6

= 11 5/6

Hence the value as a mixed fraction is 11 5/6

3) 4 3/8 + 1 5/8

= 35/8 + 13/8

= (35+13)/8

= 48/8

= 6

Hence the value is 6

4 0

3 years ago