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4 years ago

Please help me this is hard

Mathematics

2 answers:

Svetllana [295]4 years ago

6 0

10 time 8 equals 80

iren2701 [21]4 years ago

4 0

The area needs to equal 8 so you need x to be atleast 1.6. Anything over 1.6 would make the area bigger than 8 so the inequality is x>=1.6

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Please help with this math question!

Contact [7]

All you need to do is find the perimeter to see how much dense is needed
23+14+23+14=74

Answer=74 feet of fence

8 0

3 years ago

Read 2 more answers

A total of 640 tickets were sold for the school play. They were either adult tickets or student tickets. There were 60 fewer stu

gtnhenbr [62]

Answer:

350 adult tickets were sold

Step-by-step explanation:

Let x represent the number of adult tickets sold.

Since there were 60 fewer student tickets sold than adult tickets, the number of student tickets sold can be represented by x - 60

Create an equation that sets up these terms equal to 640, then solve for x:

(x) + (x - 60) = 640

2x - 60 = 640

2x = 700

x = 350

So, 350 adult tickets were sold

6 0

3 years ago

Which first step for solving the given system using substitution results in an equation wit

Paraphin [41]

I don’t know if I’m right but I got 2/3 but you should use desmos

8 0

3 years ago

According to U.S. Government regulations, a food can be labeled "lowfat" if it has at most 3 grams of fat per serving. If a norm

kaheart [24]

Determine amt of serving in 17oz
17/2=8.5 servings.
As listed above, it can have 3g per serving.
3*8.5= 25.5 grams.
The most amount of fat it can have is 25.5g. Answer is C

6 0

2 years ago

The lengths of a rectangular garden, the inner rectangle, is 9ft more than its width. For the garden. Let x = width, x + 9 = len

FinnZ [79.3K]

Answer:

$PQ = x + 13$

The dimension of the garden is 52.5ft by 43.5ft

Step-by-step explanation:

Given

The shape above

Required

Determine PQ

Determine dimension of the garden

Calculating Length PQ

Represent the length of the inner rectangle as L

Represent the width of the inner rectangle as W

$W = x$

$L = x + 9$

The distance between the inner rectangle and the outer triangle is 4ft

This implies that;

$PQ = L + 4$

Substitute $x + 9$ for $L$

$PQ = x + 9 + 4$

$PQ = x + 13$

Also;

$QR = W + 4$

Substitute $x$ for $W$

$QR = x + 4$

Calculating The Dimension of The Garden

First, we need to determine the Area of the inner rectangle

$Area_1 = L * W$

Recall that $W = x$ and $L = x + 9$

So;

$Area_1 = x * (x + 9)$

For the bigger rectangle

$Area_2 = PQ * QR$

Recall that $PQ = x + 13$ and $QR = x + 4$

So;

$Area_2 = (x + 13)(x + 4)$

Given that the Area of the walkway is $400ft^2$

This implies that

$Area_2 = Area_1 + 400$

Substitute $Area_1 = x * (x + 9)$ and $Area_2 = (x + 13)(x + 4)$

$(x + 13)(x + 4) = x * (x + 9) + 400$

Open All Brackets

$x^2 + 13x + 4x + 52 = x^2 + 9x + 400$

$x^2 + 17x + 52 = x^2 + 9x + 400$

Collect Like Terms

$x^2 - x^2 + 17x - 9x = 400- 52$

$8x = 348$

Divide both sides by 8

$\frac{8x}{8} = \frac{348}{8}$

$x = \frac{348}{8}$

$x =43.5$

Since the dimensions of the garden is $W = x$ and $L = x + 9$

<em>Substitute 43.5 for x in both cases</em>

$L = 43.5 + 9 = 52.5$

$W = 43.5$

<em>Hence, the dimension of the garden is 52.5ft by 43.5ft</em>

5 0

3 years ago