All categories

3 years ago

PLEASE PLEASE PLEASE HELP, AND SHOW WORK!!!! LOOK AT ATTACHMENT.

Mathematics

1 answer:

kaheart [24]3 years ago

5 0

Answer:

24) $230.45

Step-by-step explanation:

25%=25/100=0.25

0.25*357.35=89.38 (rounded to the nearest hundredth)

357.35-89.38=267.97

Now, we have the 25% discount, it is the same thing for the 14% discount.

14%=14/100=0.14

267.97*0.14=37.52(rounded to the nearest hundredth again)

267.97-37.52=230.45

Therefore, total cost is $230.45

You might be interested in

I need help on #5 on my summer packet my teacher never taught me this and I need help.

Drupady [299]

Answer:

Step-by-step explanation:

it can only be 0-9 that is in standard form- one digit then a decimal point

can u pls mark as brainliest?

5 0

4 years ago

Read 2 more answers

2 A cafe sells cakes and scones.

Alecsey [184]

30 cakes on tuesday

8 0

3 years ago

How do I write 6 x 3/20 as a fraction in simplest form?

Talja [164]

First, you need to turn it into a fraction by doing the multiplication. 6/1 × 3/20 which is 18/20. then you simplify it to get 9/10. 9/10 is your answer

8 0

4 years ago

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion do

V125BC [204]

Answer:

Step-by-step explanation:

It is given that

$\Delta=\begin{vmatrix}3&0&3\\2 &3&3\\0 &4&-2\end{vmatrix}$

By cofactor expansion across the first row, we get

$\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$

$\Delta=3\left[(-1)^{1+1}\begin{vmatrix}3&3\\4&-2\end{vmatrix}\right]+0\left[(-1)^{1+2}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{1+3}\begin{vmatrix}2&3\\0&4\end{vmatrix}\right]$

$\Delta=3\left[-18\right]+0\left[(-1)(-4)\right]+3\left[8\right]$

$\Delta=-54+0+24$

$\Delta=-30$

Therefore, the value of determinant is -30.

By cofactor expansion across the second column, we get

$\Delta=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}$

$\Delta=0\left[(-1)^{2+1}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{2+2}\begin{vmatrix}3&3\\0&-2\end{vmatrix}\right]+4\left[(-1)^{3+2}\begin{vmatrix}3&3\\2&3\end{vmatrix}\right]$