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ki77a [65]
3 years ago
5

Bob is mailing packages. Each small package costs him $3.40 to send. Each large package costs him $5.10. How much will it cost h

im to send 1 small package and 5 large packages?
Mathematics
1 answer:
Tatiana [17]3 years ago
8 0

Answer:

$28.9

I hope this help also I would appreciate brainliest if possible :)

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PLEASE HELP ME :( I DONT UNDERSTAND! A teacher already had a certain number of canned goods for the food drive. Each day of the
Fantom [35]
20 should be the answer because 205 divided by 10 =20
3 0
4 years ago
Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec
german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

#3

Let's start with sin here. \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}} Therefore, because sin is y/r, r = \sqrt{5} and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = \frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}

#4

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = 4\sqrt{2}. Now, to find csc, we have to do r/y.

csc = \frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

#5

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = \sqrt{7/16} = \frac{\sqrt{7}}{4}

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = -\sqrt{7/9} = \frac{-\sqrt{7}}{3}

#6

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = (\frac{-\sqrt{6}}{2})^{2}

1 + cot^2 = 6/4

cot^2 = 2/4

cot = \frac{\sqrt{2}}{2}

tan = \sqrt{2}

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

(\frac{2}{\sqrt{2}})^2 + 1 = sec^2

4/2 + 1 = sec^2

2 + 1 = sec^2

sec^2 = 3

sec = -\sqrt{3}

cos = \frac{-\sqrt{3}}{3}

Hope this helps!! :)

3 0
3 years ago
Read 2 more answers
Find the y-intercept and the slope of the line<br><br> y=-x+7<br><br> y-intercept:<br> slope:
Ipatiy [6.2K]

Answer:

The y-intercept would be 7

The slope would be -1 or -1/1. They are both the same thing.

The slope is the number which is located directly to the left of the variable (the variable in this case is x, but in other situations it could be a totally different letter).

5 0
3 years ago
If PQRS is a rhombus, find PQR.
labwork [276]
PQR is most likely a triangle
5 0
3 years ago
Read 2 more answers
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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