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Natali5045456 [20]
2 years ago
6

Find the value of x.

Mathematics
1 answer:
Novay_Z [31]2 years ago
4 0

Answer:

Step-by-step explanation:

Remark

What you don't know and what makes the problem confusing, is that where to the given arcs start and end. So I am going to say that both arcs start from E and 119 ends at D, and 174 ends at F which means that the arc FD is not known, but it is not hard to find.

The total number of degrees from beginning to end is 360 degrees. The formula for finding FD is

Formula

FD = 360 - (174 - 119)

The formula for x is (ED - FD)/2

Solution

Arc FD = 360 - (174 - 119)

Arc FD = 67

So x = (119 - 67) / 2

x = 52 /2

x = 26

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Henry bought 5/6 pound of roasted almonds for five dollars he wants to know the price per pound
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Answer:

$6

Step-by-step explanation:

p=pound

5/6p= 500

p= 500(6/5)

p=600 cents

p=$6

Hope this helps!!!

8 0
3 years ago
This is hard pls help
makkiz [27]

The length of the SM parallelogram when the length of the rectangle is 15 cm and width is 8 cm is 8/5 units.

<h3>What is the area of a rectangle?</h3>

Area of a rectangle is the product of the length of the rectangle and the width of the rectangle. It can be given as,

A=a\times b

Here, (a)is the length of the rectangle and (b) is the width of the rectangle

The length of the rectangle is 15 cm and width is 8 cm. Thus, the area of it is,

A=15\times8\\A=120\rm\; cm^2

All three parts has equal area. Thus, the area of parallelogram NCMA is,

A_p=\dfrac{120}{3}\\A_p=40\rm\; cm^2

MN is the height of the parallelogram. Thus,

A_p=h\times AS\\40=h\times15\\h=\dfrac{40}{15}\\h=\dfrac{8}{5}

Thus, the length of the Sm parallelogram when the length of the rectangle is 15 cm and width is 8 cm is 8/5 units.

Learn more about the area of rectangle here;

brainly.com/question/11202023

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Why is circle 1 similar to circle 2?
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Circle 2 is a dial at ion of circle 1
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8 0
3 years ago
Complete the table of values
Agata [3.3K]
Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify! 

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, 3^{-2} =&#10;\frac{1}{3^{2} }.
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, (\frac{3}{4}) ^{3}  =  \frac{ 3^{3} }{4^{3} }.
3) The zero exponent rule<span> says that any number raised to zero is 1. For example, 3^{0} = 1.
</span>

Back to the Problem:
Problem 1 
The x-values are in the left column. The title of the right column tells you that the function is y =  4^{-x}. The x-values are:
<span>1) x = 0
</span>Plug this into y = 4^{-x} to find letter a:
y = 4^{-x}\\&#10;y = 4^{-0}\\&#10;y = 4^{0}\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = 4^{-x} to find letter b:
y = 4^{-x}\\ &#10;y = 4^{-2}\\ &#10;y =  \frac{1}{4^{2}} \\  &#10;y= \frac{1}{16}
<span>
3) x = 4
</span>Plug this into y = 4^{-x} to find letter c:
y = 4^{-x}\\ &#10;y = 4^{-4}\\ &#10;y =  \frac{1}{4^{4}} \\  &#10;y= \frac{1}{256}
<span>

Problem 2
</span>The x-values are in the left column. The title of the right column tells you that the function is y =  (\frac{2}{3})^x. The x-values are:
<span>1) x = 0
</span>Plug this into y = (\frac{2}{3})^x to find letter d:
y = (\frac{2}{3})^x\\&#10;y = (\frac{2}{3})^0\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = (\frac{2}{3})^x to find letter e:
y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\&#10;y =  \frac{4}{9}
<span>
3) x = 4
</span>Plug this into y = (\frac{2}{3})^x to find letter f:
y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}
<span>
-------

Answers: 
a = 1
b = </span>\frac{1}{16}<span>
c = </span>\frac{1}{256}
d = 1
e = \frac{4}{9}
f = \frac{16}{81}
5 0
3 years ago
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