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3 years ago

How do i do this? please help

Mathematics

1 answer:

kherson [118]3 years ago

3 0

Y(0)=0+3=3 (0;3)
y(1)=1+3=4 (1;4)
y(2)=2+3=5 (2;5)
x+3=0
x=-3
x+3=1
x=-2
x+3=2
x=-1

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For an angle 0 with the point (-20,-21) on its terminating side, what is the value of cosine ?

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3 years ago

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4 years ago

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4 years ago

A total of $10,000 is invested in two mutual funds. The first account yields 5% and the second account yields 6%. How much was i

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Answer:

$2,500 was invested in the first account while $7,500 was invested in the second account

Step-by-step explanation:

Here in this question, we want to find the amount which was invested in each of the accounts, given their individual interest rates and the total amount that was accorded as interest from the two investments

Now, since we do not know the amount invested , we shall be representing them with variables.

Let the amount invested in the first account be $x and the amount invested in the second account be $y

Since the total amount invested is $10,000, this means that the summation of both gives $10,000

Mathematically;

x + y = 10,000 ••••••(i)

now for the $x, we have an interest rate of 5%

This mathematically translates to an interest value of 5/100 * x = 5x/100

For the $y, we have an interest rate of 6% and this mathematically translates to a value of 6/100 * y= 6y/100

The addition of both interests, gives 575

Thus mathematically;

5x/100 + 6y/100 = 575

Multiplying through by 100, we have

5x + 6y = 57500 •••••••••(ii)

From 1, we can have x = 10,000-y

let’s substitute this into equation ii

5(10,000-y) + 6y = 57500

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50,000 + y = 57500

y = 57500-50,000

y = 7,500

Recall;

x = 10,000-y

so we have;

x = 10,000-7500 = 2,500

3 0

3 years ago

QUESTION 11.1

lawyer [7]

The group paid $ 5250 at first city and $ 6250 at second city

<u>Solution:</u>

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

<em><u>Therefore, a equation is framed as:</u></em>

8.5 % of x + 5.5 % of y = 790

$\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790$

0.085x + 0.055y = 790 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

<em><u>Substitute x = 5250 in eqn 1</u></em>

y = 5250 + 1000

Thus the group paid $ 5250 at first city and $ 6250 at second city

8 0

3 years ago