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3 years ago

Keshawn is asked to compare and contrast the domain and range for the two functions.

1 answer:

5 0

For the function 5x, the range is the value of f(x) which is five times the x and the domain is the value of x which can be the ratio of f(x) and 5. For the second function which is g(x) = 5, the range is 5 all through out the graph while the domain is infinity.

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A triangle with sides 11m , 13m and 18m is a right triangle. A True B False

Julli [10]

$\bold{\huge{\pink{\underline{ Solution}}}}$

$\bold{\underline{ Given :-}}$

A triangle with sides 11m, 13m and 18m

$\bold{\underline{ To \: Find :-}}$

We have to check it whether it is right angled triangle or not?

$\bold{\underline{ Let's \: Begin :-}}$

$\sf{\red{ In \:right \:angled\ : triangle, }}$

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

We imply it in the given triangle ,

$\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}$

$\sf{(AB)² + (BC)² = (AC)²}$

$\sf{ (11)² + (13)² = (18)² }$

$\sf{ 121 + 169 ≠ 324 }$

$\sf{ 290 ≠ 324 }$

From Above we can conclude that,

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

$\sf{\blue{ Hence,\: Your\: answer \:is \:false }}$

5 0

2 years ago

There are 63 students marching in a band, and they’re marching in 7 rows. How many students are in each row?

VLD [36.1K]

Total divided by number of rows=number per row
63/7=9
9 per row

4 0

3 years ago

Read 2 more answers

Draw an area model 1.1x2.4

Alexus [3.1K]

You have to draw a rectangle and label the longer side 2.4 and the shorter side 1.1

5 0

3 years ago

Scientists are measuring the thickness of ice on a large lake. When they first measure the ice, it is 3.1 inches thick. Three we

yaroslaw [1]

Answer:

Step-by-step explanation:

What is the question?

Amount of change = 5.5-3.1 = 2.4 inches

Percentage change = 2.4/3.1 ≈ 77.4% increase

3 0

3 years ago

A 95-foot wire attached from the top of a cell phone tower makes a 62 degree angle with the ground. Joey is standing 150 feet be

a_sh-v [17]

Answer:

23.32 degrees

Step-by-step explanation:

We set up a large right triangle that has 2 triangles within it. The large triangle is a right triangle. The height of it is the height of the tower, the base angle is 62, the hypotenuse is 95, and the base measure is y. The other triangle has the same height which is the height of the tower, the angle is what we are looking for, and the base measure is 150 feet beyond y, so its measure is y + 150. We have enough information to find the height of the tower, so let's do that first. Going back to the first smaller triangle.

$sin62=\frac{x}{95}$ so the height of the tower is 83.88 feet. Now we need to solve for y. Using that same triangle and the tangent ratio, we find that $tan62=\frac{83.88}{y}$ . Now let's do the same thing for the other triangle with the unknown angle.

$tan\beta =\frac{83.88}{y+150}$

Solve both of these for y. The first one solved for y:

$y=\frac{83.88}{tan62}$

The second one solved for y will simplify to:

$y=\frac{83.88-150tan\beta }{tan\beta }$

Now that these are both solved for y, and y = y, we can set them equal to each other by the transitive property of equality:

$\frac{83.88-150tan\beta }{tan\beta }=\frac{83.88}{tan62}$

Cross multiply to get this big long messy looking thing:

$tan62(83.88-150tan\beta )=83.88tan\beta$

Distribute through the parenthesis to get

$83.88tan62-[(tan62)(150tan\beta)]=83.88tan\beta$

Get the unknown angles on the same side so it can be factored out:

$83.88tan62=83.88tan\beta +[(tan62)(150tan\beta )]$

And then factoring it out gives you:

$83.88tan62=tan\beta(83.88+150tan62)$

Divide to get

$tan\beta =\frac{83.88tan62}{83.88+150tan62}$

Do this on your calculator in degree mode to give you an angle measure of 23.32°. I know this is really hard to follow without being able to draw the pics for you like I do in my classroom, but hopefully you can follow my description and draw your own triangles and follow from that!

4 0

3 years ago