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1 year ago

Which of these is true and why

Mathematics

1 answer:

LuckyWell [14K]1 year ago

7 0

Answer:

Step-by-step explanation:

True statements

All equallateral triangles are similar. Their sides are all in the same ratio when comparred.

All squares are similar. Same reason as equilateral triangles. All sides to both squares compared are the same.

False Statements

Isosceles triangles can and usually do have different base angles.

rectangles can have all sorts or side lengths. The only requirement is consecutive sides form right angles.

2 rhombuses can have side lengths that are in the same ratio, but the heights are not in the same ratio. That eleminates.

Answer

These are the only true ones: Statements 2 and 5 are true. The rest are not.

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I GIVE BRAINIEST “What is the solution of a system of equations if the sum of the

zhenek [66]

Answer:

At the bottom!!

Step-by-step explanation:

y=2

2=0+x

x=2

Solution of a system of equations:A system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. The solution of such a system is the ordered pair that is a solution to both equations. ... The solution to the system will be in the point where the two lines intersect.

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2 years ago

If y varies directly as x, find the constant of variation if y = 25 and x = 5.

Soloha48 [4]

Answer:

The constant of variation is 5

Step-by-step explanation:

We know direct variation is of the form

y = kx

We know y =25 and x=5

25 = k*5

Divide each side by 5

25/5 = 5k/5

5 = k

3 0

2 years ago

A. f(x) = l-2xl - 3 Domain: {1, 2, 3} Find the range

gizmo_the_mogwai [7]

Hello there!

We are given the function:

$\displaystyle \large{ f(x) = | - 2x| - 3} \\$

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

x = 1

$\displaystyle \large{ f(1) = | - 2(1)| - 3} \\ \displaystyle \large{ f(1) = | - 2| - 3} \\$

Recall that any numbers in absolute value are always positive.

$\displaystyle \large{ f(1) = 2 - 3} \\ \displaystyle \large{ f(1) = - 1} \\$

x = 2

$\displaystyle \large{ f(2) = | - 2(2)| - 3} \\ \displaystyle \large{ f(2) = | - 4 | - 3} \\ \displaystyle \large{ f(2) = 4 - 3} \\ \displaystyle \large{ f(2) = 1}$

x = 3

$\displaystyle \large{ f(3) = | - 2(3)| - 3} \\ \displaystyle \large{ f(3) = | - 6| - 3} \\ \displaystyle \large{ f(3) = 6 - 3} \\ \displaystyle \large{ f(3) = 3}$

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

6 0

2 years ago

Round off 0.043 to one significant figure

aniked [119]

Hey there :)

0.043
The first 0 before the decimal point is not significant since it is used for cosmetic purpose
The 0 after the decimal is not significant as the 0 is used to locate the decimal point
The 4 is the first nonzero integer and it counts as a significant figure

Therefore, 0.04 will be your answer rounded to 1 s.f

4 0

3 years ago

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Alla [95]

Answer:

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3 years ago