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ehidna [41]
3 years ago
6

Can sum1 help me plot the point 1/4, -2 1/4, 1 1/2?

Mathematics
1 answer:
balu736 [363]3 years ago
3 0

Answer:

I do not have a graph rn but I hope I helped

Step-by-step explanation:

well, yo just have to find 0.25 on the graph, mark it then do the same for -5.25 and 5.5

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Answer:

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Prove : EG Bisects HF , HF Bisects EG

Step-by-step explanation:

Proof

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Mr liston bought lunch for $ 12.80. He gave the waiter a tip of 15 %. How much money did the waiter recieve?
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3 years ago
∆ABC is similar to ∆DEF. The ratio of the perimeter of ∆ABC to the perimeter of ∆DEF is 1 : 10. The longest side of ∆DEF measure
exis [7]

Answer:

Part 1) The length of the longest side of ∆ABC  is 4 units

Part 2) The ratio of the area of ∆ABC to the area of ∆DEF is \frac{1}{100}

Step-by-step explanation:

Part 1) Find the length of the longest side of ∆ABC

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

The ratio of its perimeters is equal to the scale factor

Let

z ----> the scale factor

x ----> the length of the longest side of ∆ABC

y ----> the length of the longest side of ∆DEF

so

z=\frac{x}{y}

we have

z=\frac{1}{10}

y=40\ units

substitute

\frac{1}{10}=\frac{x}{40}

solve for x

x=(40)\frac{1}{10}

x=4\ units

therefore

The length of the longest side of ∆ABC  is 4 units

Part 2) Find the ratio of the area of ∆ABC to the area of ∆DEF

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z ----> the scale factor

x ----> the area of ∆ABC

y ----> the area of ∆DEF

z^{2}=\frac{x}{y}

we have

z=\frac{1}{10}

so

z^2=(\frac{1}{10})^2

z^2=\frac{1}{100}

therefore

The ratio of the area of ∆ABC to the area of ∆DEF is \frac{1}{100}

3 0
3 years ago
Blain is two years older than three times Jillians age. Jillian is also 16 years younger than Blain. How old is Blain?
Ipatiy [6.2K]

The age of Blain is 23 years old

<h3><u>Solution:</u></h3>

Let the age of Blain be "a" and age of Jillian be "b"

Given that Blain is two years older than three times Jillians age

So we can frame a equation as:

age of blain = 2 + 3(age of Jillian)

a = 2 + 3b ----- eqn 1

Also given that Jillian is also 16 years younger than Blain

Age of Jillain = Age of Blain - 16

b = a - 16 ---- eqn 2

Substitute eqn 2 in eqn 1

a = 2 + 3(a - 16)

a = 2 + 3a - 48

a - 3a = -46

-2a = -46

a = 23

Thus the age of Blain is 23 years old

4 0
3 years ago
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