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34kurt
2 years ago
9

Find the unit rate. Round to the nearest hundredth, if necessary. $290 for 18 ft2

Mathematics
1 answer:
DochEvi [55]2 years ago
5 0

Answer:16.110

Step-by-step explanation:

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E<br> Homework: 1-7 HW<br> Simplify the expression.<br> 8(5+t) - 2(t+1)<br> 8(5 + t) - 2(t+1)=
lys-0071 [83]

Hi, I'm happy to help!

To simplify this expression, we need to first use the order of operations <u>(Parenthesis, Exponents, Multiplication or Division (Left to Right), and Addition or Subtraction (Left to Right)</u>), then combine like terms. (<u>80,40,57</u>, or <u>6t, 8t, 98t</u>, or<u> 2t²,7t²,83t²</u>)

When a number is next to parenthesis with no sign, it means we are using multiplication. We can change this expression to look like this, by changing all the numbers next to the parenthesis with no sign, to one with a multiplication sign.

8(5+t) - 2(t+1) 8(5 + t) - 2(t+1)=

8×(5+t) - 2×(t+1)× 8×(5 + t) - 2×(t+1)=

We start with parenthesis and solve everything inside of those, but, because everything inside the parenthesis are unlike terms, we cannot combine them.

Next, We look for any exponents (x²). There are none in this equation.

We then look for multiplication or division. There is a lot of multiplication, so we start solving that from left to right.

8×(5+t) - 2×(t+1)× 8×(5 + t) - 2×(t+1)=

40+8t - 2×(t+1)× 8×(5 + t) - 2×(t+1)=

40+8t - 2×(t+1)× 8×(5 + t) - 2×(t+1)=

(40+8t) - (2t+2)× 8×(5 + t) - 2×(t+1)=

(40+8t) - (2t+2)× 8×(5 + t) - 2×(t+1)=

(40+8t) - (16t+16)×(5 + t) - 2×(t+1)=

(40+8t) - (16t+16)×(5 + t) - 2×(t+1)=

(40+8t) - (80t+16t²+80+16t) - 2×(t+1)=

(40+8t) - (80t+16t²+80+16t) - 2×(t+1)=

(40+8t) - (80t+16t²+80+16t) - (2t+2)=

(40+8t) - (80t+16t²+80+16t) - (2t+2)=

Now, before we do anything else, we need to recheck the parenthesis, now that they have changed. We can further simplify the middle parenthesis by adding together like terms.

(40+8t) - (80t+16t²+80+16t) - (2t+2)=

(40+8t) - ((80t+16t)+16t²+80) - (2t+2)=

(40+8t) - (96t+16t²+80) - (2t+2)=

(40+8t) - (96t+16t²+80) - (2t+2)=

Now, we look for addition or subtraction of like terms. We can now remove the parenthesis, since everything inside of the last two parenthesis is negative, we turn it negative, as long as we only combine the like terms.

40+8t - 96t-16t²-80 - 2t-2=

Once we find like terms, we can rearrange the expression.

(40-80-2)+8t - 96t+16t² - 2t

(-42)+8t - 96t-16t² - 2t

(-42)+8t - 96t-16t² - 2t

(-42)+(8t-96t-2t)-16t²

(-42)+(-90t)-16t²

(-42)+(-90t)-16t²

This doesn't have any like terms, so we don't alter it.

(122)+(-90t)-16t²

We can now remove the parenthesis.

122-90t-16t²

Since there are no more like terms this is the furthest we can simplify. We can then rearrange the expression to be arranged completely correct.

-16t²-90t-42

I hope this was helpful, keep learning! :D

4 0
3 years ago
PLEASE CAN SOMEONE HELPPP
NikAS [45]

Answer:

second option

Step-by-step explanation:

The perimeter P is the sum of the 3 sides, then

P = 4x² + 3 + 2x - 5 + x + 11 ← collect like terms

= 4x² + 3x + 9

4 0
3 years ago
Please help select the equivalent representations of the interval written below
zhuklara [117]

Answer:

3rd option

Step-by-step explanation:

x < 7

means that x cannot equal 7 but all values less than 7

since not equal at either end of interval use parenthesis, not brackets , then

(- ∞ , 7 ) ← is the required interval

6 0
1 year ago
2m=p-q/r solve for r
Oduvanchick [21]

To solve for r, you would start by subtracting p from both sides.

2m - p = -q/r

Multiply everything by r.

2mr - pr = -q

Factor r out of the left side of the equation.

r(2m - p) = -q

Divide both sides by (2m - p).

r = \frac{-q}{2m-p} = \frac{q}{2m + p}.

6 0
3 years ago
Make n the subject of the formula
kogti [31]

Answer:

\mathsf{ \frac{m + 21}{5} = n }

Step-by-step explanation:

To make "n" the subject of the formula, rearrange the formula so it begins with " n = "

To isolate the variable "n", you need to inverse the other terms on that side of the equation where "n" really is.

=> m = 5n - 21

  • <em>there's</em><em> </em><em>a</em><em> </em><em>"</em><em>-21</em><em>"</em><em> </em><em>next</em><em> </em><em>to</em><em> </em><em>"</em><em>5n</em><em>"</em><em>,</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>subtraction</em><em> </em><em>is</em><em> </em><em>addition</em><em>,</em><em> </em><em>so</em><em> </em><em>add</em><em> </em><em>21</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em>.</em><em> </em>

=> m + 21 = 5n <u>-</u> <u>21</u> + <u>21</u>

=> m+ 21 = 5n

  • <em>There's</em><em> </em><em>also</em><em> </em><em>a</em><em> </em><em>5</em><em> </em><em>attached</em><em> </em><em>to</em><em> </em><em>n</em><em>,</em><em> </em><em>that</em><em> </em><em>means</em><em> </em><em>the</em><em> </em><em>multiplication</em><em> </em><em>of</em><em> </em><em>n</em><em> </em><em>with</em><em> </em><em>5</em><em>,</em><em> </em><em>the</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>multiplication</em><em> </em><em>is</em><em> </em><em>division</em><em>,</em><em> </em><em>so</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em> </em><em>by</em><em> </em><em>5</em>

<em>\mathsf{ \frac{m + 21}{5} = n }</em>

3 0
3 years ago
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