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spayn [35]
2 years ago
10

Which statement is true? (table included and answers)

Mathematics
1 answer:
Masteriza [31]2 years ago
5 0

Answer:

b

Step-by-step explanation:

i think the answers b

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1. Charlie bought a new computer for $450 with his credit card. He plans to pay the minimum payment
atroni [7]

Given:

Charlie bought a new computer for $450 with his credit card.

He plans to pay the minimum payment  of $25 per month.

To find:

The rate of change.

Solution:

Let y be the remaining amount of loan after x months.

Initial amount of loan = $450

Payment of one month = $25

Payment of x months = $25x

Now,

Remining amount = Initial amount of loan - Payment of x months

y=450-25x     ...(i)

The slope intercept form is

y=mx+b     ...(ii)

where, m is slope and b is y-intercept.

From (i) and (ii), we get

m=-25

Therefore, the rate of change is -25 dollars.

8 0
3 years ago
How do you do these questions?
Alika [10]

Answer:

a) - 7 dollars

b) 24.30 dollars

c) - 5 dollars

Step-by-step explanation:

b) 27 / 10%

Divide 27 by 10/100 by multiplying 27 by the reciprocal of 10/100.

27 x 100/10

Multiply 27 and 100 to get 2700.

2700/10

Divide 2700 by 10 to get 270.

270

Divide 270 by 100 to get 2.7

Subtract 27.00 - 2.7 = 24.30

24.30

4 0
2 years ago
I need help with this plz
Morgarella [4.7K]

Answer:

i would go with B

Step-by-step explanation:

I'm not too sure I'm so sorry but thats my closest guess

8 0
2 years ago
Can I get some help ???
Nikitich [7]

Answer is D. <em><u>288</u></em> . i am glad to help

4 0
2 years ago
A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
2 years ago
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