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3 years ago

How to get 100 by multiplying 12

Mathematics

2 answers:

expeople1 [14]3 years ago

8 0

12 x 8.33 = 100 (100 ÷ 12 = 8.33)

Reptile [31]3 years ago

5 0

You multiply 12x8.3 and get the answer 100

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Find the slope of the line passing through the points (-5,1) and (2,4).

Leno4ka [110]

Answer:

<h3> $\boxed{ \bold{ \huge{ \boxed{ \sf{ \frac{3}{7} }}}}}$ </h3>

Step-by-step explanation:

Let the points be A and B

A ( -5 , 1 ) ⇒( x₁ , y₁ )

B ( 2 , 4 )⇒( x₂ , y₂ )

Finding the slope of the line passing through these Points

Slope = $\sf{ \frac{y2 - y1}{x2 - x1} }$

plug the values

⇒ $\sf{ \frac{4 - 1}{2 - ( - 5)} }$

We know that $\sf{( - ) \times ( - ) = ( + )}$

⇒ $\sf{ \frac{4 - 1}{2 + 5} }$

Subtract 1 from 4

⇒ $\sf{ \frac{3}{2 + 5} }$

Add the numbers

⇒ $\sf{ \frac{3}{7} }$

Hope I helped!

Best regards!!

7 0

2 years ago

1.3(8-b)+3.7b= -5.2 solve and check answer

Paraphin [41]

Answer:

b = -6.5

Step-by-step explanation:

I am 100% sure thts the right answer

6 0

3 years ago

Read 2 more answers

What is the cost of a $19,500 plus a 6% sales tax?

vova2212 [387]

1% of 19,500= 195

6% = 1%x6 = 195x6 = 1170

19500+1170=20670

3 0

3 years ago

The percentage of sprouted seeds for a single crop can be modeled by a logistic function that is represented by y=100/1+99e^-0.8

olga_2 [115]

Answer:

A, B, and C

Step-by-step explanation:

8 0

3 years ago

Jill bought a car for $50,000. Her car depreciates at a rate of 10% a year. create an equation for the problem

Vadim26 [7]

Answer:

The value of car after n years at the depreciation rate is $ 50,000 $(0.9)^{n}$ .

Step-by-step explanation:

Given as :

The cost of the car that Jill bought = $ 50,000

The depreciation rate of car value = r = 10 % a years

Let The car after n years of depreciation = $ A

Now, According to question

The cost of car after n years of depreciation = initial cost of car × $(1 - \dfrac{\textrm rate}{100})^{\textrm time}$

Or, $ A = $50,000 × $(1 - \dfrac{\textrm r}{100})^{\textrm n}$

Or, $ A = $50,000 × $(1 - \dfrac{\textrm 10}{100})^{\textrm n}$

Or, $ A = $50,000 × $(\frac{90}{100})^{n}$

I.e $ A = $50,000 × $(0.9)^{n}$

So, value of car after n years = $ 50,000 $(0.9)^{n}$

Hence The value of car after n years at the depreciation rate is $ 50,000 $(0.9)^{n}$ . Answer

6 0

2 years ago