All categories

2 years ago

What is the measure of the angle at the tail end of the kite? if the top area is where its pointed is 122?

Mathematics

1 answer:

nataly862011 [7]2 years ago

8 0

Answer:

The angle is 58 degrees

Step-by-step explanation:

Given

See attachment for kite

Required

The angle at the tail end

Represent this angle with x.

From the attached kite, we have:

1 angle = 122

2 angles = right-angled

So, we have:

$x + 122 +90+90 = 360$ --- sum of angles in a kite

$x + 302 = 360$

Solve for x

$x =- 302 + 360$

$x =58^\circ$

You might be interested in

Don’t get it please helppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

UNO [17]

What do you don’t get

7 0

3 years ago

A manatee surfaced for air 6 times in 240 seconds. How many seconds went by before the manatee surfaced the fourth time if it he

Norma-Jean [14]

I think it is 160 seconds

3 0

2 years ago

Describe how to transform:

Anna71 [15]

Answer:

$x^\frac{35}{6}$

Step-by-step explanation:

The expression to transform is:

$(\sqrt[6]{x^5})^7$

Let's work first on the inside of the parenthesis.

Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:

$\sqrt[n]{a} = a^{\frac{1}{n}}$

Therefore $\sqrt[6]{a} = a^{\frac{1}{6}}$

Now let's replace $a$ with $x^{5}$ which is the algebraic form we are given inside the 6th root:

$\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}$

Now use the property that tells us how to proceed when we have "exponent of an exponent":

$(a^n)^m= a^{n*m}$

Therefore we get: $(x^5)^{\frac{1}{6}}=x^{\frac{5}{6}}}$

Finally remember that this expression was raised to the power 7, therefore:

$[tex](\sqrt[6]{x^5})^7=(x^\frac{5}{6})^7=x^\frac{35}{6}$ [/tex]

An use again the property for the exponent of a exponent:

8 0

2 years ago

Please help this is due tomorrow.<br><br> 1/2 of 20?

Mars2501 [29]

1/2 of 20 is 20/2 which is 10.

All you really need to do here is to divide 20 by 2 to get 10.

6 0

2 years ago

Read 2 more answers

How many of the numbers from 10 through 92 have the sum of their digits equal to a perfect square?

horrorfan [7]

All the numbers in this range can be written as $10d_1+d_0$ with $d_1\in\{1,2,\ldots,9\}$ and $d_2\in\{0,1,\ldots,9\}$ . Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)

so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.

For each number that occupies an entire diagonal in the table, it's easy to see that that number $n$ shows up $n$ times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.

So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.

7 0

3 years ago