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3 years ago

Solve for b. 75 - 15 = 56 - 3 b =

Mathematics

1 answer:

ANEK [815]3 years ago

4 0

Answer:4/3

Step-by-step explanation:

75-15-56=3b

4=3b

b=4/3

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Lily bought 22.17 pounds of grapefruit. The lightest grapefruit weighed 1.3 pounds. The heaviest grapefruit weighed 1.7 pounds.

nalin [4]

Answer:

Approximately 15 grapefruits

Step-by-step explanation:

Average Value

The mean or average of a sample is a number that represents the whole set by the center value of them. The formula to compute the average is

$\displaystyle \bar x=\frac{\sum x_i}{n}$

Where xi is the value of the element i and n is the number of elements in the set

We need to find an estimate of how many grapefruits Lily purchased knowing the heaviest and the lightest weight of the set. We can find the average value by using the two extreme values:

$\displaystyle \bar x=\frac{1.3+1.7}{2}=1.5$

If each grapefruit weighed 1.5 pounds, then we can estimate the 22.17 pounds of them will contain

$22.17/1.5\approx 15\ \text{grapefruits}$

3 0

3 years ago

a local newspaper story reports that 8% of the 7968 students at the college work full time, how many students work full time

e-lub [12.9K]

It will be (7968*8)/100=637.44

3 0

3 years ago

Please help me guys ill mark you as a brilliant

Tomtit [17]

Answer:

so the answer is ,,5". hope this helped! Have a nice day.

Step-by-step explanation:

45:9=5

4 0

3 years ago

Find the imaginary part of\[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]

iren [92.7K]

Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

$x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6$

First, we can expand this power using the binomial theorem:

$(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}$

After that, we can apply De Moivre's theorem to expand each summand: $(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)$

The final step is to find the common factor of i in the last expansion. Now:

$x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6$

$=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6$

$=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))$

The last part is to multiply these factors and extract the imaginary part. This computation gives:

$Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288$

$Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288$

(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)

A calculator simplifies the imaginary part Im(x⁶) to 0

4 0

3 years ago

X+5 3 + x−5 2x+5 = x−5 3x

Lina20 [59]

Answer: 4x+5 $\neq$ 4x-5

Step-by-step explanation: It's really hard to tell what the question is.

8 0

3 years ago