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2 years ago

Write the following expression as a single logarithm with coefficient 1. log910 − log9 1 2 − log94

Mathematics

2 answers:

sesenic [268]2 years ago

8 0

Answer:

its a

Step-by-step explanation:

nikitadnepr [17]2 years ago

6 0

The answer is A. Log 9 5

I got it right on ed .

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Saira weights 25 pounds more than umbar . If together they weight 205 pounds. What is the weight of saira

goblinko [34]

To solve this, you can create an equation. Say the weight of Umbar is x, then Saira weighs x+25. Together they weigh 205. So, you get the equation

x+x+25=205

Then solve for x.

2x+25=205
2x=180
x=90

Then to find Saira's weight, add 25 to Umbar's weight.

90+25=115.

6 0

2 years ago

n a right triangle, the tangent of angle A is 912 and the cosine of angle A is 1215. Based on these values, what is the sine of

Allisa [31]

Answer:

Option A $sin(A)=\frac{9}{15}$

Step-by-step explanation:

we know that

$tan(A)=sin(A)/cos(A)$

Solve for sin(A)

$sin(A)=tan(A)cos(A)$

we have

$tan(A)=\frac{9}{12}$

$cos(A)=\frac{12}{15}$

substitute

$sin(A)=\frac{9}{12}(\frac{12}{15})$

$sin(A)=\frac{9}{15}$

5 0

2 years ago

2. Starting at a fixed time, each car entering an intersection is observed to see whether it turns left (L), right (R), or goes

jasenka [17]

Answer:

Natural numbers (integers greater than zero)

X = 3, 5, 4, 4, 3

Step-by-step explanation:

The least number of cars that can be observed in this experiment is 1, if the first car turns left. On the other hand, the experiment could go on forever if no car ever turns left, thus the highest number of cars approaches infinite.

The possible values of X are integers greater than zero, which are known as the Natural numbers.

If X = number of cars observed, simply count the number of letters in each outcome for the value of X:

Outcome = RRL, AARRL, AARL, RRAL, ARL

X = 3, 5, 4, 4, 3

7 0

2 years ago

The length of the day in Boulder (Latitude 40 N) can be modeled approximately by l(t) = −3 cos ( 2π 365 (t + 10)) + 12 where l i

Harman [31]

Answer:

(a) I(355)=9

(b) I'(265)=−0.05163

Step-by-step explanation:

The given function is

$I(t)=-3\cos (\frac{2\pi}{365}(t+10))+12$

where I(t) the length of the day in Boulder and l is given in hours and t is the day of the year.

(a)

Substitute t=355 in given function.

$I(355)=-3\cos (\frac{2\pi}{365}(355+10))+12$

$I(355)=-3\cos (2\pi)+12$

$I(355)=-3(1)+12$

$I(355)=9$

Therefore, the value of I(355) is 9. It means the length of the day in Boulder is 9 hours at 355 day of the year.

(b)

Differentiate the given function with respect to t.

$I'(t)=-3(-\sin (\frac{2\pi}{365}(t+10)))(\frac{2\pi}{365})$

$I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(t+10))$

Substitute t=265 in the above function.

$I'(t)=\frac{6\pi}{365}\sin (\frac{2\pi}{365}(265+10))$

$I'(265)\approx −0.05163$

Therefore, the value of I'(265) is −0.05163. It means the length of the day in Boulder decreased by 0.05163 hours at 265 day of the year.

(c)

Differentiate the I'(t) function with respect to t.

$I''(t)=\frac{6\pi}{365}\cos (\frac{2\pi}{365}(t+10))(\frac{2\pi}{365})$

$I''(t)=\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))$

Equate $I''(t)=0$ to find critical points.

$\frac{12\pi^2}{(365)^2}\cos (\frac{2\pi}{365}(t+10))=0$

$\cos (\frac{2\pi}{365}(t+10))=0$

$\cos (\frac{2\pi}{365}(t+10))=\cos (\frac{\pi}{2})$

On comparing both sides we get

$\frac{2\pi}{365}(t+10)=\frac{\pi}{2}$

$t+10=\frac{\pi}{2}\times \frac{365}{2\pi}$

$t+10=\frac{365}{4}$

$t=\frac{365}{4}-10$

$t=81.25$

Check the value of the function I'''(t) at t=81.25.

Since $I'''(81.25)=-0.0000153$ , therefore the value of l′(t) is largest at 81.25.

4 0

2 years ago

Subtract 4 ounces from 5 pounds.

malfutka [58]

Answer is 4 lb.,12 oz. As one pound has 16 ounces. When we convert the given 5 lb. into ounces we multiply it by 16. 5 lb = 5 x 16 oz 5 lb = 80 oz When we subtract 4 oz from 80 oz We get 80 - 4 = 76 oz Which is 64 oz + 12 oz (16 x 4) oz + 12 oz Which will be 4 lb. + 12 oz So the answer will be 4 lb., 12 oz.

6 0

2 years ago