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3 years ago

PART A:

Mathematics

1 answer:

Ann [662]3 years ago

4 0

Part A:
9x12=108 felines a year
Part B:
24 felines
bonus :
84 sold

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How many pairs of whole number have a sum of 110

Andre45 [30]

2*55

5*22

10*11

1*110 are the numbers I came up with

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4 years ago

The hour hand of a clock moves -30 every hour. How many degrees does it move in two and one fifth hours?

gregori [183]

About -17.27 degrees

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3 years ago

NEED HELP ASAP!!!!

Bezzdna [24]

600g - 210d + -0.032= 15yr or c

7 0

3 years ago

A candy store sells caramels and milk chocolate by the pound the table below shows the total cost in dollars for a pound of each

MakcuM [25]

The cost of $1\frac{3}{4}$ pounds chocolate milk is $6.25 more than the cost of $1\frac{3}{4}$ pounds of Caramel.

Step-by-step explanation:

Given,

Cost of one pound of Caramels = $10.28

Cost of $1\frac{3}{4}$ pounds of Caramel = $10.28*1\frac{3}{4}$

Cost of $1\frac{3}{4}$ pounds = $10.28*\frac{7}{4}=\frac{71.96}{4}= \$17.99$

Cost of one pound of chocolate milk = $13.85

Cost of $1\frac{3}{4}$ pounds = $13.85*1\frac{3}{4}=13.85*\frac{7}{4}$

Cost of $1\frac{3}{4}$ pounds = $\frac{96.95}{4}=\$24.24$

Difference = 24.24 - 17.99 = $6.25

The cost of $1\frac{3}{4}$ pounds chocolate milk is $6.25 more than the cost of $1\frac{3}{4}$ pounds of Caramel.

Keywords: difference, multiplication

Learn more about multiplication at:

#LearnwithBrainly

5 0

3 years ago

1. Simplify. 5–1(3–2)

konstantin123 [22]

Answer with explanation:

Using following law of indices

$1.x^a \times x^b=x^{a+b}\\\\2. (x^m)^n=x^{mn}\\\\ 3. x^{-n}=\frac{1}{x^n}\\\\4. \frac{x^m}{x^n}=x^{m-n}$

$1.5^{-1}\times 3^{-2}=\frac{1}{5}\times\frac{1}{3^2}\\\\=\frac{1}{5\times9}\\\\=\frac{1}{45}\\\\2. \frac{mn^{-4}}{p^0*q^{-2}}=\frac{mq^2}{n^4}\\\\ 3.0.0042=4.2\times 10^3\\\\4.6.12 \times 10^3=6.12 \times 1000=6120\\\\ 5.0.5 \times(8\times10^5})=\frac{1}{2}\times8 \times 10^5=4\times 10^5\\\\6. (9 \times 10^3)^2=9^2 \times (10^3)^2=81 \times 10^6\\\\7.( \frac{1}{2}a^{-4}\times b^2 )_{a=-2,b=4}{\text{or}}( \frac{1}{2}a^{4}\times b^{-2} )_{a=-2,b=4}=\frac{1}{2}\times (-2)^{-4}\times(4)^2{\text{or}}=\frac{1}{2}\times (-2)^{4}\times(4)^{-2}=\frac{16}{2\times16}=\frac{1}{2}$

$9. (4\times x\times y^2)^3\times (xy)^5=4^3\times x^3\times (y^2)^3\times x^5 \times y^5\\\\=64\times x^{3+5}\times y^{5+6}\\\\=64x^8y^{11}\\\\10.(2\times t^3)^3(0.4\times r)^2=2^3 \times (t^3)^3\times (0.4)^2 \times r^2=8\times t^9 \times 0.16 \times r^2=1.28r^2t^9$