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mixas84 [53]
2 years ago
13

Evaluate the expression when x = 12/7

Mathematics
1 answer:
lyudmila [28]2 years ago
6 0

Answer:

82

Step-by-step explanation:

1/3( x+9/7) + 3^4

Let x = 12/7

1/3( 12/7+9/7) + 3^4

PEMDAS says parentheses first

1/3( 21/7) + 3^4

1/3(3) +3^4

Then exponents

1/3(3)+81

Then multiply

1+81

82

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What is the probability of picking a blue coin out of a bag that contains 5 blue and 10 red coins
Neko [114]

the probability is 5/15 or 1/3.

You add up all the coins (5+10=15) and since there’s 5 blue coins in the bag, you put 5 over 15. You simplify that to 1/3.

hope this helped !!

6 0
3 years ago
The area of a triangle is 195 square centimeters. The length of the base is 26 centimeters. What is the height of the triangle?
Natali5045456 [20]
To find the area of a triangle uses the equation base*height/2. For this equation, we have 26 * h/2 = 195. This means that 26 * h = 195 * 2.
26 * 15 = 390, so your answer is 15 cm.
7 0
3 years ago
Help! Please? The ratio (by volume) of milk in a certain solution is 3 to 8. If the total volume of the solution is 187 cubic fe
DIA [1.3K]
We know that the total volume of the solution is 187 cubic feet.
187=3x+8x
187=11x
17=x
Now we know that the ratio 3 to 8 is 51 to 136.
Double check, (3*17)+(8*17)=187.
I'm not sure which part of the ratio is water, but knowing this information, you should be able to solve it. I believe it may be this:
The total volume of milk is 51. If the rest of the solution is water, then 187-51=136.
6 0
3 years ago
A)A cuboid with a square x cm and height 2xcm². Given total surface area of the cuboid is 129.6cm² and x increased at 0.01cms-¹.
Nutka1998 [239]

Answer: (given assumed typo corrections)


(V ∘ X)'(t) = 0.06(0.01t+3.6)^2 cm^3/sec.


The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.



Part B) y'(x+Δx/2)×Δx gives exactly the same as y(x+Δx)-y(x), 0.3808, since y is quadratic in x so y' is linear in x.


Step-by-step explanation:

This problem has typos. Assuming:

Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.


129.6 cm^2 = 2(base cm^2) + 4(side cm^2)

= 2(X cm)^2 + 4(X cm)(2X cm)

= (2X^2 + 8X^2)cm^2

= 10X^2 cm^2

X^2 cm^2 = 129.6/10 = 12.96 cm^2

X cm = √12.96 cm = 3.6 cm


so X(t) = (0.01cm/sec)(t sec) + 3.6 cm, or, omitting units,

X(t) = 0.01t + 3.6

= the length parameter after t seconds, in cm.


V(X) = 2X^3 cm^3

= the volume when the length parameter is X.


dV(X(t))/dt = (dV(X)/dX)(X(t)) × dX(t)/dt

that is, (V ∘ X)'(t) = V'(X(t)) × X'(t) chain rule


V'(X) = 6X^2 cm^3/cm

= the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).


X'(t) = 0.01 cm/sec

= the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.

V(X(t)) = (V ∘ X)(t) = 2(0.01t+3.6)^3 cm^3

= the volume after t seconds, in cm^3

V'(X(t)) = 6(0.01t+3.6)^2 cm^2

= the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.

(V ∘ X)'(t) = ( 6(0.01t+3.6)^2 cm^3/cm )(0.01 cm/sec) = 0.06(0.01t+3.6)^2 cm^3/sec

= the rate of change of the volume per change in time, in cm^3/sec, after t seconds.


Problem to ponder: why is (V ∘ X)'(t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?


Question part b)


Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.


This is a little ambiguous, but "use differentiation" suggests that we want y'(4.02) yunit per xunit, rather than Δy/Δx = (y(4.02)-y(4))/(0.02).


Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y(x) at all.

Then we want y'(x+Δx/2)×Δx


y(x) = 2x^2 + 3x

y'(x) = 4x + 3


y(4) = 44

y(4.02) = 44.3808

Δy = 0.3808

Δy/Δx = (0.3808)/(0.02) = 19.04


y'(4) = 19

y'(4.01) = 19.04

y'(4.02) = 19.08


Estimate Δy = (y(x+Δx)-y(x)/Δx without evaluating y() at all, using only y'(x), given x = 4, Δx = 0.02.


y'(x+Δx/2)×Δx = y'(4.01)×0.02 = 19.04×0.02 = 0.3808.


In this case, where y is quadratic in x, this method gives Δy exactly.

6 0
3 years ago
Gisselle has 1/2 of a candy bar. She role off 1/2 of what she had. She then took the piece she broke off and split it into three
saul85 [17]

Answer:

\frac{1}{12}

Step-by-step explanation:

Given:

Gisselle has 1/2 of a candy bar.

She role off 1/2 of what she had and split it into three equal pieces.

Question asked:

What fraction of a candy bar was each piece?

Solution:

Gisselle has = \frac{1}{2} \ of \ a \ candy \ bar

She role off = \frac{1}{2} \ of \ what\ she \ had

                     =\frac{1}{2}  \ of\  \frac{1}{2}\\=\frac{1}{2} \times \frac{1}{2}\\\\=\frac{1}{4}

As she split it into three equal pieces, we will divided it by 3, we get,

\frac{1}{4} \div3\\\frac{1}{4}\times\frac{1}{3}

\frac{1}{12}

Therefore, fraction of a candy bar of each piece is \frac{1}{12}

4 0
3 years ago
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