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3 years ago

Multiply and answer on scientific notation : (3.8 x 10^8) (6.9 x 10^5)

Mathematics

2 answers:

Kruka [31]3 years ago

4 0

(3.8 x 10^8) (6.9 x 10^5) = 2.622e+14

dexar [7]3 years ago

3 0

(3.8×10^8)×(6.9×10^5)= 2.622e^14

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What isn’t the minimum value of c=7x+8y given the constraints on x and y listed below

Luda [366]

Answer:

32 is not minimum value of C

Step-by-step explanation:

Given constraints:

$2x+y\geq8$

$x+y\geq8$

$x\geq8$

$y\geq8$

Now we draw the graph of each constraint and see the feasible region in graph. Please have a look attached graph.

We got three corner points A(0,8) B(2,4) and C(6,0)

Now we find the optimal solution for each value of corner points.

C(0,8)=7(0)+8(8)=64

C(2,4)=7(2)+8(4)=14+32=46

C(6,0)=7(6)+8(0)=42

Option A is correct. 32 is not minimum value of C

7 0

4 years ago

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A cube has a side length s of 7 centimeters. The expression 6s2 gives the surface area of the cube and the expression s3 gives t

JulijaS [17]

Answer:

Area = 294 cm^2

Volume = 343 cm^3

Step-by-step explanation:

Simply plug in s = 7 into the given equations.

$6s^{2} = 6*7^{2} = 294 cm^{2}$

$s^{3} = 7^{3} = 343 cm^{3}$

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2 years ago

onsider the equation 0.5 times e4z =13 solve the equation for z. express the solution as a logarithm in base e

olga_2 [115]

Answer:

$z=\frac{ln(26)}{4}$

Step-by-step explanation:

The given equation is:

$0.5\,e^{4z}=13$

so, we first isolate the exponential form that contains the unknown "z" in the exponent, by dividing both sides by 0.5:

$0.5\,e^{4z}=13\\e^{4z}=\frac{13}{0.5}\\e^{4z}=26$

Now we bring the exponent down by applying the natural log function on both sides, and then solve for "z":

$e^{4z}=26\\ln(e^{4z})=ln(26)\\4\,z=ln(26)\\z=\frac{ln(26)}{4}$

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4 years ago

What is the common ratio for this geometric sequence?

Anastasy [175]

Answer:

Step-by-step explanation:

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3 years ago

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Use the empirical rule. the mean speed of a sample of vehicles along a stretch of highway is 66 miles per hour, with a standard

Romashka-Z-Leto [24]

I dont know this one i am very sorry

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3 years ago

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