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2 years ago

Can somebody answer this please? Will give brainlest :) (valid answers only or you’ll be reported)

Mathematics

2 answers:

vova2212 [387]2 years ago

7 0

Answer:

The slope of the line is 0.

Step-by-step explanation:

To find the slope of a line, you need to find their rise over run. I will be using the two points (1, 4) and (5, 4).

$\frac{rise}{run}$ = $\frac{4-4}{5-1}$

= $\frac{0}{4}$

= 0

atroni [7]2 years ago

3 0

The line does not go up or go down, it stays straight, therefore the slope is 0.

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The scale on a map is 55 cm: 88 km. If the distance between two cities is 5656 km, how far apart in cm are the two cities on the

Alinara [238K]

Answer:

3,535 cm

Step-by-step explanation:

55cm=88km

Let's make that "one unit" - 55cm=1 unit. How many units are in the distance 5656km?

5656km/88km=64.2727... units.

55cmx64.2727... = 3,534.999 cm (depending on what your teacher requires, we can probably round to 3,535).

Therefore, we know that the two cities are 3,535 cm apart on the map.

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2 years ago

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Ymorist [56]

The answer is 63.6 degrees Fahrenheit or rounded up it wpuld be 64 degrees

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3 years ago

a miniature railway increased its prices by 25% to £1.50 how much did it cost to visit the railway before

MAVERICK [17]

The answer is 112.5

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3 years ago

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1. Bob tried to answer the following question by finding the missing angle and rounding the answer to the nearest degree.

iren2701 [21]

Answer:

Part 1)

Bob's mistake was to have used the cosine instead of the sine

The measure of the missing angle is $53.13\°$

Part 2) The surface area of the pyramid is $288\ cm^{2}$

Step-by-step explanation:

Part 1)

Let

x----> the missing angle

we know that

In the right triangle o the figure

The sine of angle x is equal to divide the opposite side angle x to the hypotenuse of the right triangle

$sin(x)=\frac{16}{20}$

$x=arcsin(\frac{16}{20})=53.13\°$

Bob's mistake was to have used the cosine instead of the sine

Part 2) we know that

The surface area of the square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces

$SA=b^{2}+4[\frac{1}{2}(b)(h)]$

where

b is the length side of the square

h is the height of the triangular lateral face

In this problem

$h=b/2$ -------> by an 45° angle

$b=2h$

$sin(45\°)=\frac{h}{6\sqrt{2}}$

$h=6\sqrt{2}(sin(45\°))=6\ cm$

Find the value of b

$b=2(6)=12\ cm$

Find the surface area

$SA=12^{2}+4[\frac{1}{2}(12)(6)]=288\ cm^{2}$

3 0

3 years ago

What is the average rate of change for the line shown in the graph

kompoz [17]

Answer:

The average rate of change is 10.

Step-by-step explanation:

Since the average rate of change is the slope of a line, you'd solve this problem by finding the slope.

First, you can start by using the $\frac{y_2-y_1}{x_2-x_1}$ method.

Plug in the points (0,10) and (1,20).

$\frac{20-10}{1-0} =\frac{10}{1} =10$

Therefore your answer is 10.

Hope this helped!

6 0

2 years ago