All categories

3 years ago

Explain how drawing triangles can determine the sum of the interior angles of the quadrilateral?

Mathematics

1 answer:

Kobotan [32]3 years ago

3 0

Answer: each triangle drawing is 180 degrees total, 2 triangles is 360 etc.

Step-by-step explanation:

You might be interested in

The graph of an exponential function passes through (2,−32) and (3,−64). Find the exponential function that describes the graph.

Greeley [361]

Answer: $y=(-8)2^x$

Step-by-step explanation:

The exponential function will have this form:

$y=ab^x$

We know that the function passes through the points $(2,-32)$ and $(3,-64)$ . Then, we can substitute the coordinates of the point $(2,-32)$ into $y=ab^x$ and solve for "a":

$-32=ab^2\\\\a=\frac{-32}{b^2}$

Then, we know that:

$y=(\frac{-32}{b^2})b^x$

Now, we neeed to substitute the coordinates of the second point $(3,-64)$ into $y=(\frac{-32}{b^2})b^x$ and solve for "b":

$-64=(\frac{-32}{b^2})b^3\\\\-64=-32b\\\\\frac{-64}{-32}=b\\\\b=2$

Substituting the value of "b" into $a=\frac{-32}{b^2}$ we can find "a":

$a\frac{-32}{2^2}\\\\a=-8$

Therefore, we get that the exponential function that describes the graph, is:

$y=(-8)2^x$

7 0

3 years ago

A group of 6 friends want to share 8 pears evenly. All of the pears are the same size. How many pears should each friend receive

konstantin123 [22]

1.3333333 pears each

4 0

3 years ago

Read 2 more answers

URGENT HELP PLZ MATH!

7nadin3 [17]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Angles in a triangle total to 108 degrees

180 - (45 + 15) = 120

Angle B is 120 degrees.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0

3 years ago

Read 2 more answers

Find the exact value of the expression.

Tresset [83]

$\sin(a-b)=\sin a \cos b-\cos a \sin b$ . If we let $a=\cos^{-1} \left(\frac{5}{6} \right)$ and $b=\tan^{-1} \left(\frac{1}{2} \right)$ , then the given expression is equal to:

$\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)$

Using the Pythagorean identities $\sin^{2} x+\cos^{2} x=1$ and $\tan^{2} x+1=\sec^{2} x$ ,

$1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}$

$2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

$\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

This means we can write the original expression as:

$\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}$

6 0

2 years ago

Read 2 more answers

Simplify this radical.<br> X^13<br> O 13/х<br> O 6xX<br> 0 xVx12<br> 0 xox

Mademuasel [1]

Solución de problemas

6x 2x3xx

6 0

2 years ago