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3 years ago

A 6-pack of light bulbs costs $7.38. What is the unit price?

Mathematics

1 answer:

Alika [10]3 years ago

7 0

Answer:

ever beer cost 1.38 so that is it

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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

I will try to give as many details as possible.

First of all, I just would like to say:

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Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! （＊＾Ｕ＾）人（≧Ｖ≦＊）/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

Solve the inequality -1/3h ≥ 7 A. h ≥ –21 B. h ≤ 7 1/3 C. h ≤ –21 D. h ≥ -2 1/3

JulsSmile [24]

The Answer Is C Fam...................

8 0

3 years ago

Read 2 more answers

Which coordnates represent a point on the graph of 3y = 5x - 13?

Elena L [17]

(x,y)
sub given points and see if true

A, (0,-13)
x=0
y=-13
3(-13)=5(0)-13
-39=0-13
-39=-13
false (other user was just guessing)

B. (3,1)
x=3,
y=1
3(1)=5(3)-13
3=15-13
3=2
false

C. (7,7)
x=7
y=7
3(7)=5(7)-13
21=35-13
21=22
false

D. (-6,-1)
3(-1)=5(-6)-13
-3=-30-13
-3=-43
false

answer is none of them

3 0

3 years ago

Work out the value of angle x and how ??

frutty [35]

Answer:

160 degrees

Step-by-step explanation:

Step 1: A straight line is 180 degrees. So 180 - 100, is 80 degrees. as the opposite angle measurement of the 100 degree.

Step 2: An isosceles triangle has two equal angle measurements, so two of its angles are 80 degrees.

Step 3: All angles in a triangle equal 180 degrees. So add them up (we will call the missing angle, Z) 180 = 80+80+Z. which equals 180= 160+z

Step 4: Solve it. You subtract 160 from both side which comes out to 20 = Z.

Step 5: Now you have the opposite angle of X. Going back to step 1, A straight angle is 180 degrees. 180 - 20 = 160. X = 160 degrees

6 0

2 years ago

Given the linear programming problem, use the method of corners to determine where the minimum occurs and give the minimum value

algol [13]

Answer:

Minimum value of function $C=x+10y$ is 63 occurs at point (3,6).

Step-by-step explanation:

To minimize :

$C=x+10y$

Subject to constraints:

$x\leq 3---(1)\\y\leq 9---(2)\\x+y\geq 9----(3)\\x\geq 0\\y\geq 0$

Eq (1) is in blue in figure attached and region satisfying (1) is on left of blue line

Eq (2) is in green in figure attached and region satisfying (2) is below the green line

Considering $x+y\geq 9$ , corresponding coordinates point to draw line are (0,9) and (9,0).

Eq (3) makes line in orange in figure attached and region satisfying (3) is above the orange line

Feasible region is in triangle ABC with common points A(0,9), B(3,9) and C(3,6)

Now calculate the value of function to be minimized at each of these points.

$C=x+10y$

at A(0,9)

$C=0+10(9)\\C=90$

at B(3,9)

$C=3+10(9)\\C=93$

at C(3,6)

$C=3+10(6)\\C=63$

Minimum value of function $C=x+10y$ is 63 occurs at point C (3,6).

3 0

3 years ago