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Sergio039 [100]
3 years ago
12

Peter spent half the money on his gift card and coffee and he loaded another $10 on the gift card how much was on the gift card

to begin with if he has spent $40 on the card
Mathematics
1 answer:
cestrela7 [59]3 years ago
6 0

Answer:

$80 was on the gift card to begin with

Step-by-step explanation:

Let the initial money be x

Peter spent (1/2)*x

Peter added 10

<em>Before adding more money on the gift card Peter had x -(1/2)x left on the gift card which is (1/2)x</em>

<em>After loading more money Peter had (1/2)x + 10</em>

(1/2)x = 40 <em>(money spent by Peter)</em>

x/2 = 40

x = 80

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Mehvish buys 1/2 kg carrots, 3 kg potatoes, 2 kg onions and 1 1/2kg of chicken.
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Answer:

The amount of change will be £7.99

Step-by-step explanation:

Here, we want to calculate the amount of change that Mehvish should get

1/2 kg carrots at 50 p per kg will be;

1/2 * 50 = 25 p

3 kg potatoes at 64 p per kg will be

3 * 64 = p 192

2 kg onions at 42 p per kg

2 * 42 p = 84 p

1 1/2 kg chicken at £6 per kg will be:

since £1 is 100 p

1.5 * 6 = £9 (9 * 100 p = 900p)

since £1 is 100 p

£20 will be;

20 * 100 = 2,000 p

So let us make the subtraction as follows;

2,000- 900 - 84-192 - 25 = 799 p

In pounds, this is divided by 100

we have this as 799/100 = £7.99

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3 years ago
What is 7% tax on a $9.79 purchase
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Will give brainiest answer. Which pair of ratios does NOT form a proportion?
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3 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!! :)

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3 years ago
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