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3 years ago

Find the missing term in the pattern.

Mathematics

2 answers:

KatRina [158]3 years ago

8 0

Answer:

Step-by-step explanation:

the pattern is increase the number by one more than the increase before. so 6,8=2 greater

8-11=3 greater. 11-15=4 greater. so, 15+5=20 (with this answer being 5 greater continuing the pattern.)

Whitepunk [10]3 years ago

7 0

Answer:

Step-by-step explanation:

6 + 2 = 8

8 + 3 = 11

11 + 4 =15

15 + 5 =20

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A die is rolled find the probability of rolling a multipul of 5

Pavlova-9 [17]

Answer:

1/6

Step-by-step explanation:

It's 1/6 because the only multiple of 5 in a die is 5, so it will be 1/6.

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3 years ago

Can i have help please. Step by step

nignag [31]

Answer: It's in the step-by step explanation

Step-by-step explanation:

I just learned about this too. I'll use what I know to help you out.

According to whatever law of the circle, where you have two intersecting lines within the bounds of a circle(that'd be TQ and SW), the product of the divided segments will equal each other.

So to put that in terms, TU times QU = SU times WU.

So let's get the value of segment TU, which is 1.5

Then let's get the value of segment of QU, which is 4.

Now let's get the value of WU, which is 3. We don't know what SU is yet. So put it in algebraic form.

1.5(4)=3x

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3 years ago

Read 2 more answers

Plss answer!! <br>MERRY Christmas!!! <br>Critical Question. AAAAA

liberstina [14]

Answer:

Step-by-step explanation:

abc = 1

We have to prove that,

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

We take left hand side of the given equation and solve it,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$

Since, abc = 1,

$\frac{1}{c}=ab$ and c = $\frac{1}{ab}$

By substituting these values in the expression,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+ab}+\frac{1}{1+\frac{1}{ab}+\frac{1}{a}}$

$=\frac{b}{b+ab+1}+\frac{1}{1+b+ab}+\frac{ab}{ab+1+b}$

$=\frac{1+b+ab}{1+b+ab}$

$=1$

Which equal to the right hand side of the equation.

Hence, $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

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