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mamaluj [8]
3 years ago
13

H(m)+42= need help with problem​

Mathematics
1 answer:
kondaur [170]3 years ago
7 0
It depends on H(m) what is it?
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Carol wants to make a cake. The recipe calls for 8 1/2 cups of flour. A 16-ounce bag contains 2 cups. How many bags of flour mus
Leona [35]

Answer:

please love me

Step-by-step explanation:

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3 years ago
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Rowan reads at a constant rate of 1.5 pages per minute and has already
saul85 [17]

Using linear functions, it is found that Rowan and Kennedy will have read the same number of pages in 10 minutes.

What is a linear function?

  • A linear function is modeled by:

y = ax + b

In which:

  • a is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0.

For Rowan, we have that:

  • He has already read 20 pages, hence the y-intercept is b = 20.
  • He reads 1.5 pages per minute, hence the slope is a = 1.5

Then:

y_R = 1.5x + 20

For Kennedy, we have that:

  • He has already read 25 pages, hence the y-intercept is b = 25.
  • He reads 1 page per minute, hence the slope is a = 1

Then:

y_K = x + 25

They will have read the same number of pages after x minutes, for which:

y_R = y_K

Then:

1.5x + 20 = x + 25

0.5x = 5

x = \frac{5}{0.5}

x = 10

To learn more about linear function, you can take a look at brainly.com/question/25823744

3 0
3 years ago
What is the solution to the problem 2y+3=4y+2
yulyashka [42]

Answer:

y = 1/2

Step-by-step explanation:

<u>Step 1:  Solve for y</u>

2y + 3 = 4y + 2

2y + 3 - 2 - 2y = 4y + 2 - 2 - 2y

1 / 2 = 2y / 2

1/2 = y

Answer:  y = 1/2

8 0
3 years ago
Read 2 more answers
You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the
Otrada [13]

Answer:

Please check the explanation.

Step-by-step explanation:

a)  How much will you have at the middle of the first year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

Total amount = A = ?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

substituting the values

A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}

A=300\cdot \frac{2.06}{2}

A=\frac{618}{2}

A=309 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

Total amount = A = ?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

so substituting the values

A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}

A=300\cdot \frac{2.06^2}{2^2}

A=318.27 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

7 0
3 years ago
Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the pairs of polynomials to their products.
andreyandreev [35.5K]

The products of the polynomials are:

  • (xy + 9y + 2) * (xy - 3) = x²y² - xy + 9xy² - 27y - 6
  • (2xy + x + y) * (3xy² - y) = 6x²y³ - 2xy² + 3x²y² -xy + 3xy³- y²
  • (x - y) * (x + 3y) = x² + 2xy + 3y²
  • (xy + 3x + 2) * (xy – 9)  = x²y² - 7xy + 3x²y - 27x  - 18
  • (x² + 3xy - 2) * (xy + 3)  = x³y + 3x² + 3x²y² + 7xy - 6
  • (x + 3y) * (x – 3y) = x² - 9y²

<h3>How to evaluate the products?</h3>

To do this, we multiply each pair of polynomial as follows:

<u>Pair 1: (xy + 9y + 2) and (xy – 3)</u>

(xy + 9y + 2) * (xy - 3)

Expand

(xy + 9y + 2) * (xy - 3) = x²y² - 3xy + 9xy² - 27y + 2xy - 6

Evaluate the like terms

(xy + 9y + 2) * (xy - 3) = x²y² - xy + 9xy² - 27y - 6

<u>Pair 2: (2xy + x + y) and (3xy² - y)</u>

(2xy + x + y) * (3xy² - y)

Expand

(2xy + x + y) * (3xy² - y) = 6x²y³ - 2xy² + 3x²y² -xy + 3xy³- y²

<u>Pair 3: (x – y) and (x + 3y) </u>

(x - y) * (x + 3y)

Expand

(x - y) * (x + 3y) = x² + 3xy - yx + 3y²

Evaluate the like terms

(x - y) * (x + 3y) = x² + 2xy + 3y²

<u>Pair 4: (xy + 3x + 2) and (xy – 9) </u>

(xy + 3x + 2) * (xy – 9)

Expand

(xy + 3x + 2) * (xy – 9)  = x²y² - 9xy + 3x²y - 27x + 2xy - 18

Evaluate the like terms

(xy + 3x + 2) * (xy – 9)  = x²y² - 7xy + 3x²y - 27x  - 18

<u>Pair 5: (x² + 3xy - 2) and (xy + 3) </u>

(x² + 3xy - 2) * (xy + 3)

Expand

(x² + 3xy - 2) * (xy + 3)  = x³y + 3x² + 3x²y² + 9xy - 2xy - 6

Evaluate the like terms

(x² + 3xy - 2) * (xy + 3)  = x³y + 3x² + 3x²y² + 7xy - 6

<u>Pair 6: (x + 3y) and (x – 3y)</u>

(x + 3y) * (x – 3y)

Apply the difference of two squares

(x + 3y) * (x – 3y) = x² - 9y²

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

5 0
2 years ago
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