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3 years ago

5

I need help with this please help the last one is 20 for 4.20

1 answer:

Anuta_ua [19.1K]3 years ago

5 0

Hello!

The word "per" means to divide. Therefore we will divide our price by ounces to find how much one ounce costs.

.42/2=.21

It costs $0.21 per ounce.

I hope this helps!

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Simplify the following expression into the form a + bi, where a and b are rational numbers.

vagabundo [1.1K]

The answer is -16 - 10i.

Using the distributive property on the first part, we have:
-2i*7--2i*4i + (3+i)(-2+2i)
-14i+8i² +(3+i)(-2+2i)

Using FOIL on the last part,
-14i+8i²+(3*-2+3*2i+i*-2+i*2i)
-14i+8i²-6+6i-2i+2i²
-10i+8i²-6+2i²

Since we know that i = -1,
-10i+8(-1)-6+2(-1)
-10i-8-6-2
-16-10i

7 0

3 years ago

Read 2 more answers

(55 - 52 ) + (+3 + 6)<br> Order of operations

True [87]

Answer: 12

Step-by-step explanation:

(55-52) + ( 3 +6)

3 + 9

12

6 0

3 years ago

Use the properties of real numbers to rewrite the expression. 2/5 * (-7) * 5/2

Mama L [17]

For this case we have the following expression:

$\frac{2}{5}*(- 7)*\frac{5}{2}$

Using the associative property we can rewrite the expression in the following way:

$(\frac{2}{5}*\frac{5}{2})*(-7)$

Finally, simplifying we have:

$(1) (- 7)$

$-7$

Answer:

Rewriting the expression we have that the result is:

d. -7

4 0

3 years ago

All 6 members of the Valastro family have decided to

atroni [7]

Answer:

5

Step-by-step explanation:

7 0

3 years ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

3 years ago