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Rzqust [24]
3 years ago
14

A takeaway sells 10-inch pizzas and 12-inch pizzas.

Mathematics
1 answer:
N76 [4]3 years ago
5 0

Answer:

532.52

Step-by-step explanation:

P 2(10-inch pizza)=407*3.72-358*0.49=1514.04-175.42=1339.62

P2(12-inch pizza)=169*5.26-142*0.04=888.94-5.68=883.26

P=1339.62+883.26=2222.88

2222.88-1690.36=532.52

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3 years ago
-4x + 7 = -21 <br><br>whats x AHSAJ I NEEDD HELPPP
yKpoI14uk [10]

Answer:

-4x+7= -21

-4x= -21-7( this positive 7 will change to negative as we take in other side )

-4x = -28

Both the negative will be cancelled

4x = 28

x = 28/4

X = 7

therefore the X value is 7

7 0
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Which of the following do indexes and scales have in common?
statuscvo [17]

Answer:

c.They rank-order the units of analysis in terms of specific variables

Step-by-step explanation:

An index is a way  of compiling one score from a variety of  questions or statement that represents a belief, feeling or attitude.

Scales on the other hand measure levels of intensity at a variable level like how much a person agrees or disagrees with a particular statement.

The one thing common to both  indexes and scales is that

They rank-order the units of analysis in terms of specific variables.

4 0
3 years ago
Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi
stealth61 [152]
Applying our power rule gets us our first derivative,

\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}

simplifying a little bit,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

looking for critical points,

\rm 0=2x^{-2/3}+4x^{1/3}

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

0=2x^{-2/3}\left(1+2x\right)

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

\rm f'(x)=2x^{-2/3}+4x^{1/3}

and take our second derivative.

\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}

Looking for inflection points,

\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}

Again, pulling out the smaller power of x, and fractional part,

\rm 0=-\frac43x^{-5/3}\left(1-x\right)

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
8 0
3 years ago
Trying to see what my answer would be
Oksanka [162]

Answer:

B. 2^{-14}*5^{10}.

Step-by-step explanation:

Use the power of a power property.

(2^{-7}*5^{5} )^{2}

Multiply all the exponents in the parenthesis by 2.

(2^{-7}*5^{5} )^{2}=(2^{-14}*5^{10} )

The answer is B. 2^{-14}*5^{10}.

Hope this helps!

7 0
2 years ago
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