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2 years ago

The rule g = b + 3 expresses the relationship between the

Mathematics

2 answers:

Pani-rosa [81]2 years ago

6 0

Answer:

g = 3 + b

b + 3 = g

3 + b = g

Step-by-step explanation:

jeka57 [31]2 years ago

3 0

Answer:

[see below]

Step-by-step explanation:

Rewriting the rule so 'b' would be the subject:

$g = b + 3\\\\g - 3 = b + 3-3\\\\g-3=b\\\\\boxed{b=g-3}$

Hope this helps.

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The green darner is one of the largest dragonflies. It weighs about four hundredths of an ounce. Write four hundredths in standa

Tju [1.3M]

Answer:

Four hundredths

In standard form - $4\times 10^{-2}$

Step-by-step explanation:

Given : The green darner is one of the largest dragonflies.

It weighs about four hundredths of an ounce.

To write : Four hundredths in standard form.

Solution :

Standard form is when we write a larger number in a smaller way.

For example : $20000= 2\times 10^4$

So, according to question

Largest dragonflies weighs about four hundredths of an ounce.

Four hundredths means 0.04

In standard form - $4\times 10^{-2}$

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2 years ago

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1.1 gallons hope u the best good luck

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2 years ago

Jarrod helped his teachers pack textbooks into boxes for the summer. He packed 7 boxes of science textbooks with 14 books per bo

serious [3.7K]

Answer:

Step-by-step explanation:

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3 years ago

70° 13 ४ c= 12.2 x A

Aleonysh [2.5K]

Answer:

sin70=13/x

$\sin(70) \times x = 13$

$x = 13 \times \sin(70)$

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2 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

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3 years ago