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Sunny_sXe [5.5K]

2 years ago

14

Can u plz help me i have 2 more to day after this one

2 answers:

gavmur [86]2 years ago

7 0

$\frac{5 {}^{2} \times {5}^{6} }{5 {}^{4} } \\ { \frac{5}{5 {}^{4} } }^{2 + 6} \\ \frac{ {5}^{8} }{5 {}^{4} } \\ 5 {}^{8 - 4} \\ 5 {}^{4}$

so 5⁴ is the answer

Shalnov [3]2 years ago

6 0

Answer:

$?=4$

Step-by-step explanation:

$\frac{5^2*5^6}{5^4}=5^?\\ \\\frac{5^{2+6}}{5^4}=5^?\\ \\\frac{5^8}{5^4}=5^?\\\\5^{8-4}=5^?\\\\5^4=5^?\\\\?=4$

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A 46 gram sample of a substance that is used to sterilize surgical instruments has a k-value of 0.1374. Find the substance's hal

mars1129 [50]

Answer:

$t=5.0days$

Step-by-step explanation:

Using the formula for the exponential decay that is $N=N_{0}e^{-kt}$ , we have N= $\frac{1}{2}{\times}46=23$ , $N_{0}=46$ and k=0.1374.

Thus, $N=N_{0}e^{-kt}$ becomes

$23=46{\times}e^{-0.1374t}$

$\frac{23}{46}=e^{-0.1374t}$

$\frac{1}{2}=e^{-0.1374t}$

Taking log on both sides, we get

$ln(\frac{1}{2})={-0.1374t}$

$t=\frac{ln\frac{1}{2}}{-0.1}$

$t=\frac{-0.6931}{-0.1374}$

$t=5.0days$

4 0

3 years ago

Two quantities, Pand Q, are connected by

frez [133]

Answer:

P = 0.5Q - 20

Step-by-step explanation:

P = kQ + c

substitute Q = 60 , P = 10 into the equation

10 = 60k + c → (1)

substitute Q = 240, P = 100 into the equation

100 = 240k + c → (2)

subtract (1) from (2) term by term to eliminate c

90 = 180k ( divide both sides by 180 )

$\frac{90}{180}$ = $\frac{1}{2}$ = 0.5 = k

substitute k = 0.5 into (1) and solve for c

10 = 60(0.5) + c

10 = 30 + c ( subtract 30 from both sides )

- 20 = c

then

P = 0.5Q - 20

8 0

2 years ago

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

4 0

3 years ago

Find the value of x for which the lines a and b are parallel. answer : 23 20 2 19

Afina-wow [57]

Answer:

x=19

Step-by-step explanation:

These two angles are alternate interior angles and alternate interior angles are equal when the lines are parallel

148 = 7x+15

Subtract 15 from each side

148-15 =7x+15-15

133 = 7x

Divide by 7

133/7 = 7x/7

19 =x

4 0

3 years ago

A square has a side length of 12 cm. Which of the following is closest to the length of its diagonal? A) 9 cm B) 12 cm C) 12v 2

kondaur [170]

C) 12V2
12^2 + 12^2 = c^2
144 + 144 = 288
V288 = 12V2

6 0

2 years ago