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2 years ago

The ratio of the lengths of 3 squares of different sizes is 2:4:7. If the length of the biggest

Mathematics

1 answer:

Slav-nsk [51]2 years ago

8 0

9514 1404 393

Answer:

Step-by-step explanation:

The largest square is 7-2 = 5 ratio units larger than the smaller square. Hence, each ratio unit stands for (10 cm)/5 = 2 cm. The side length of the smaller square is 2(2 cm) = 4 cm. The perimeter is 4 times that: 16 cm.

The perimeter of the smallest square is 16 cm.

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2. Choose the graph that shows the solution of the inequality on the number line

horsena [70]

Answer:

Number line 3 also known as [C]

Step-by-step explanation:

Given that:

c ≤ -1

Knowing that:

≤ = Less than or equal to

Now substitute into the equation:

C less than or equal to -1

Thus, it going left on a number line.

The only number line with left is [A] and [C]

Since ≤ = Less than or equal to -1

then, it including -1

also known as a closed circle means the number is included and an open circle means it is not.

Hence, it closed circle.

Thus, the answer is [C]

Kavinsky

4 0

1 year ago

Write 321x64 in long muliplication

makkiz [27]

Step by step:
3 2 1
× 6 4
————————————-
+ 1 2 8 4
+ 1 9 2 6 0
————————————-
= 2 0 5 4 4

4 0

2 years ago

What is the probability that something with a 2.18% chance of occurring happens 3 times out of 194 events

k0ka [10]

Answer:

0.18431525 = 18.4%

Step-by-step explanation:

General Formula :

total trials Cn⋅p(success)^n⋅p(fail)^total−n

1198144* (.0218)^3 * (1-.0218)^191

4 0

2 years ago

Prove the following by induction. In each case, n is apositive integer. 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

To prove: $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

The formula for finding the perimeter of a rectangle is p= 2l 2w solve the formula for w.

I am Lyosha [343]

Answer:

w = $\frac{p -2l}{2}$

Step-by-step explanation:

p = 2l + 2w Subtract 2l from both sides of the equation

p - 2l = 2p Divide both sides by 2

$\frac{p -2l}{2}$ = w

5 0

1 year ago