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2 years ago

Of 3,000 pets owners surveyed,40% purchased their pets from a breeder.how many pets were purchased from the breeder

Mathematics

2 answers:

lord [1]2 years ago

6 0

3000 * .40 = 1200

1200 purchased from a breeder

vladimir1956 [14]2 years ago

6 0

1,200 pets were purchased from the breeder. You find this by multiplying 3,000 by 0.40 (40%)

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a bag contains 60 marbles, some blue and some green. the ratio of blue marbles to green ones is 1:5. how many blue marbles are t

olga55 [171]

Hello!

For every 1 blue marble there are 5 greens

You add the numbers in the ratio

1 + 5 = 6

Divide this number from the amount of marbles

60 / 6 = 10

Multiply the ratio by this

1 * 10 = 10

5 * 10 = 50

There are 10 blue marbles and 50 green ones

The answer is 10

Hope this helps!

4 0

2 years ago

Find the number of rectangles in 8×8 chessboard. (not just the 1x1 squares)

Gekata [30.6K]

Answer:

288

Step-by-step explanation:

Number of rectangles

____________________________________________________________

1 × 1 squares 9-1 = 1 = 8*8 = 64

1 × 2 squares 9-2 = 7 = 8*7 = 56

1 × 3 squares 9-3 = 6 = 8*6 = 48

1 × 4 squares 9-4 = 5 = 8*5 = 40

1 × 5 squares 9-5 = 4 = 8*4 =32

1 × 6 squares 9-6 = 3 = 8*3 = 24

1 × 7 squares 9-7 = 2 = 8*2 = 16

1 × 8 squares 9-8 = 1 = 8*1 = 8

64 + 56 + 48 + 40 + 32 + 24 + 16 + 8 = 288

Thus,

the number of squares on a chessboard is 228.

Hope this helps :)

8 0

2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$