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3 years ago

13

Jazziel bought a smart phone 5 years ago for $90. His sister paid 30% more for the same model. Show your work in the space provi

ded to answer the following questions. Part A: How much more did his sister pay for her phone? a) $27 b) $30 c) $36

1 answer:

Ratling [72]3 years ago

3 0

Answer:

B

Step-by-step explanation:

90/3=30

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Mikaela purchased a paddleboard originally priced at $699. If all merchandise was on sale for 40% off and the state tax rate is

vitfil [10]

Answer:

299.87

Step-by-step explanation:

$699*0.40=279.60\\279.60*0.0725= 20.271 (round)\\279.60+20.27=299.87$

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2 years ago

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Calculate the average rate of change for the function f(x)=x^4+3x^3-5x^2+2x-2, from x=-1 to x=1

borishaifa [10]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)= x^4+3x^3-5x^2+2x-2 \qquad 
\begin{cases}
x_1=-1\\
x_2=1
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\\\\\\
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6 0

3 years ago

At the beginning of football season, Coach Carnes takes inventory of the team equipment to see what he needs. He counts 24 footb

Leviafan [203]

Answer:

13

Step-by-step explanation:

To solve this, lets first find the number of footballs he still needs:

75-24 = 51

Now to find the number of packages, divide by 4, since there are 4 balls per package:

51/4 = 12 Remainder 3

Since there is a remainder, and the Coach MUST have at least 75, then we need to add a package to include the remainder:

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3 years ago

Y=x + 4 and 2y = -2x - 8

Sphinxa [80]

Socratic is a really good app for these type of questions! I definitely recommend. For science too!

5 0

3 years ago

Suppose that A = PDP-1 . Prove that det(A) = det(D)

daser333 [38]

Answer:

Check.

Step-by-step explanation:

To prove it we need to know that for two matrices A and B we have that:

det(AB) = det(A)*det(B) and $det(A^{-1}) = \frac{1}{det(A)}$ . Now:

$A = PDP^{-1}$

$det(A) = det(PDP^{-1})$

$det(A) = det(P)*det(D)*det(P^{-1})$

$det(A) = det(P)*det(D)*\frac{1}{det(P)}$

$det(A) = det(P)*\frac{1}{det(P)}*det(D)$

$det(A) = det(D)$ .

5 0

3 years ago