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3 years ago

Rule: Add 3, subtract 2. first term 7

2 answers:

6 0

By adding 3 then subtracting 2, you're pretty much just adding 1.

Starting from 7;

7, 8, 9, 10, 11, 12, etc.

I hope that helps!

Ulleksa [173]3 years ago

5 0

It is easier to just simplify that, so yu would get 1 (if 3-2)
And, just add 1 to the next number: 8, 9, 10, 11...

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

The ratio of students to adults on a field trip is 8 to 1. Which table correctly shows this ratio for each grade?

elena-14-01-66 [18.8K]

Answer:

The table C correctly shows the ratio 8:1 for each grade

Step-by-step explanation:

Let

x ----> the number of students

y ----> the number of adults

we know that

$\frac{x}{y}=\frac{8}{1}$

Verify each table

Table A

grade 6

$\frac{96}{88}=\frac{8}{1}$

Multiply in cross

$96(1)=8(88)$ ----> is not true

Table B

grade 6

$\frac{96}{104}=\frac{8}{1}$

Multiply in cross

$96(1)=8(104)$ ----> is not true

Table C

grade 6

\frac{96}{12}=\frac{8}{1}

Multiply in cross

$96(1)=8(12)$

$96=96$ ----> is true

grade 7

$\frac{120}{15}=\frac{8}{1}$

Multiply in cross

$120(1)=8(15)$

$120=120$ ----> is true

grade 8

\frac{136}{17}=\frac{8}{1}

Multiply in cross

$136(1)=8(17)$

$136=136$ ----> is true

therefore

The table C correctly shows the ratio 8:1 for each grade

Table D

grade 6

$\frac{96}{11}=\frac{8}{1}$

Multiply in cross

$96(1)=8(11)$ ----> is not true

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3 years ago

Read 2 more answers

A bowling alley has fixed cost and charges a variable per game rate. It costs $20.50 for five games and $28.50 for nine games

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Answer:

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Step-by-step explanation:

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3 years ago

3.1 In the sketch below ABC is not drawn to scale. Construct ABC next to its rough sketch below.

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Answer:

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3 years ago

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What is the exact value of sin(theta + beta)?

NISA [10]

Answer: $\frac{-3\sqrt{13}+4\sqrt{3}}{20}$

This is the single fraction of -3*sqrt(13)+4*sqrt(3) up top all over 20.

sqrt = square root

=======================================================

Explanation:

Angle theta is between pi and 3pi/2. This places the angle in quadrant Q3 where both cosine and sine are negative

Use the pythagorean trig identity to get the following:

$\sin^2 \theta + \cos^2 \theta = 1\\\\\sin^2 \theta + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1\\\\\sin^2 \theta + \frac{3}{16} = 1\\\\\sin^2 \theta = 1 - \frac{3}{16}\\\\\sin^2 \theta = \frac{16}{16} - \frac{3}{16}\\\\\sin^2 \theta = \frac{16-3}{16}\\\\\sin^2 \theta = \frac{13}{16}\\\\\sin \theta = -\sqrt{\frac{13}{16}} \ \text{ ... sine is negative in Q3}\\\\\sin \theta = -\frac{\sqrt{13}}{\sqrt{16}}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\$

Angle beta is in Q1 where sine and cosine are positive.

Draw a right triangle with legs 3 and 4. The hypotenuse is 5 through the pythagorean theorem. In other words, we have a 3-4-5 right triangle.

Since $\tan \beta = \frac{3}{4}$ , this means $\sin \beta = \frac{3}{5} \ \text{ and } \ \cos \beta = \frac{4}{5}$

Use these ideas:

sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent

In this case we have: opposite = 3, adjacent = 4, hypotenuse = 5.

-------------------------------------

To recap:

$\cos \theta = -\frac{\sqrt{3}}{4}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\\cos \beta = \frac{3}{5}\\\\\sin \beta = \frac{4}{5}\\\\$

They lead to this

$\sin\left(\theta + \beta\right) = \sin \theta * \cos \beta - \cos \theta * \sin \beta\\\\\sin\left(\theta + \beta\right) = -\frac{\sqrt{13}}{4} * \frac{3}{5} - \left(-\frac{\sqrt{3}}{4}\right) * \frac{4}{5}\\\\\sin\left(\theta + \beta\right) = -\frac{3\sqrt{13}}{20}+\frac{4\sqrt{3}}{20}\\\\\sin\left(\theta + \beta\right) = \frac{-3\sqrt{13}+4\sqrt{3}}{20}\\\\$

6 0

2 years ago