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Wittaler [7]
3 years ago
6

I wear my snow boots If and only if it snows.

Mathematics
1 answer:
hoa [83]3 years ago
6 0
Yes u only we're it if itsnows
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What is 10 2/1 divided by 2 1/4
stepladder [879]

Answer:

5.3 but the three is repeated

Step-by-step explanation:

6 0
3 years ago
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Jerry bought a coat on sale for 30% off. The sale price for the coat was $63, what is the regular price for the coat?
NNADVOKAT [17]

Answer:

The original regular price for the coat is $90

Step-by-step explanation:

The given information in the question are;

The percentage that was off (removed) from the sale price = 30%

The sale price at which Jerry bought the coat = $63

Whereby the regular price = P, we have;

P - 30% of P = $63

P - 0.3×P = $63

P×(1 - 0.3) = $63

0.7·P = $63

P = $63/0.7 = $90

Therefore, the regular (original) price for the coat = $90

The coat regular price at which the coat is displayed is at $90.

4 0
3 years ago
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 5 − x 2 . What are the dimensions
kherson [118]

Answer:

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?

Width =

Height =

Width =√10 and Height = \frac{10}{4}

Step-by-step explanation:

Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)

are (h,k) and (-h,k).

Hence, the area of the rectangle will be (h + h) × k

Therefore, A = h²k ..... (2).

Now, from equation (1) we can write k = 5 - h² ....... (3)

So, from equation (2), we can write

A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}

For, A to be greatest ,

\frac{dA}{dh} =0 = 10h-4h^{3}

⇒ h[10-4h^{2} ]=0

⇒ h^{2} =\frac{10}{4} {Since, h≠ 0}

⇒ h = ±\frac{\sqrt{10} }{2}

Therefore, from equation (3), k = 5 - h²

⇒ k=5-\frac{10}{4} =\frac{10}{4}

Hence,

Width = 2h =√10 and

Height = k =\frac{10}{4}.

8 0
3 years ago
A flat rectangular piece of aluminum has a perimeter of 62 inches. The length is 15
Salsk061 [2.6K]

Answer:

So, the required width of rectangular piece of aluminium is 8 inches

Step-by-step explanation:

We are given:

Perimeter of rectangular piece of aluminium = 62 inches

Let width of rectangular piece of aluminium = w

and length of rectangular piece of aluminium  = w+15

We need to find width i.e value of x

The formula for finding perimeter of rectangle is: Perimeter=2(Length+Width)\\

Now, Putting values in formula for finding Width w:

Perimeter=2(Length+Width)\\\\62=2(w+15+w)\\62=2(2w+15)\\62=4w+30\\62-30=4w\\4w=32\\w=\frac{32}{4}\\w=8

After solving we get the width of rectangular piece :w = 8

So, the required width of rectangular piece of aluminium is 8 inches

6 0
3 years ago
Help and I’ll kiss u again mwah mwah
ElenaW [278]

Answer:

It was reflected down and to the right I think.

Step-by-step explanation:

8 0
3 years ago
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