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3 years ago

For one minute Every 5th caller will win a Music player and every 7th caller

Mathematics

1 answer:

Assoli18 [71]3 years ago

8 0

to do this you have to find the lcm of 5 & 7 which is 35

-hope this helps

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7- a. What is the radius of the circle with center (3,10) that passes through (12,12)?

loris [4]

Answer: a. Radius of circle = $\sqrt{85}$

b. The equation of this circle : $(x-3)^2+(y-10)^2=85$

Step-by-step explanation:

Given : Center of the circle = (3,10)

Circle is passing through (12,12).

a. To find the radius we apply distance formula (∵ Radius is the distance from center to any point ion the circle.)

Radius of circle = $\sqrt{(12-3)^2+(12-10)^2}$

Radius of circle = $\sqrt{(9)^2+(2)^2}=\sqrt{81+4}=\sqrt{85}$

i.e. Radius of circle = $\sqrt{85}$

b. Equation of a circle = $(x-h)^2+(y-k)^2=r^2$ , where (h,k)=Center and r=radius of the circle.

Put the values of (h,k)= (3,10) and r= $\sqrt{85}$ , we get

$(x-3)^2+(y-10)^2=(\sqrt{85})^2$

$(x-3)^2+(y-10)^2=85$

∴ The equation of this circle : $(x-3)^2+(y-10)^2=85$

6 0

3 years ago

Martha earns $18.60 per hour. Joe earns

Soloha48 [4]

Answer:

$18.60 as well

Step-by-step explanation:

5 0

3 years ago

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Quantitative data

wariber [46]

The answer is A) measures things using quantities or numbers. Hope this helps!

8 0

3 years ago

2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar

tia_tia [17]

9514 1404 393

Answer:

a) E = 6500 -50d

b) 5000 kWh

c) the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

Given:

starting excess (E): 6500 kWh

usage: 50 kWh/day (d)

Find:

a) E(d)

b) E(30)

c) E(150)

Solution:

a) The exces is linearly decreasing with the number of days, so we have ...

E(d) = 6500 -50d

b) After 30 days, the excess remaining is ...

E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

c) After 150 days, the excess remaining would be ...

E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0

3 years ago

Simplify the expression below and write it as a single logarithm:

OLEGan [10]

The simplification of 3log(x + 4) – 2log(x – 7) + 5log(x - 2) - log(x^2) is $\log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

Solution:

Given, expression is $3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

We have to write in as single logarithm by simplifying it.

Now, take the given expression.

$\rightarrow 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

Rearranging the terms we get,

$\left.\rightarrow 3 \log (x+4)+5 \log (x-2)-2 \log (x-7)+\log \left(x^{2}\right)\right)$

$\text { since a } \times \log b=\log \left(b^{a}\right)$

$\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)$

$\text { We know that } \log a \times \log b=\log a b$

$\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.$

$\text { We know that } \log a-\log b=\log \frac{a}{b}$

$\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

Hence, the simplified form $\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

4 0

3 years ago

Read 2 more answers