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2 years ago

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The total length of the stage is 76 feet. If the trap doors are centered across the stage, what is the distance from the left si

de of the stage to the first trap door?
Please help me step by step :)

1 answer:

weqwewe [10]2 years ago

5 0

subtract the 26 1/2 from 76 and then divide by 2

76-26 1/2 = 49 1/2 / 2 = 24 3/4

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The area of a rectangle is 6x^2-9x-15. And the length is (2x-5). Which expression best represents the rectangle’s width?

tester [92]

<h2><u>AnswEr :</u> </h2>

The area of the rectangle is of the form 6x² - 9x - 15

Factoring the given expression,

$\sf \: {6x}^{2} - 9x - 15 \\ \\ \leadsto \: \sf \: 3\bigg( {2x}^{2} - 3x - 5 \bigg) \\ \\ \leadsto \: \sf \: 3 \bigg( {2x}^{2} - 5x + 2x - 5 \bigg) \\ \\ \leadsto \: \sf \: 3 \bigg((2x - 5)(x + 1) \bigg)$

Length of the rectangle is 2x - 5

Thus,

$\leadsto \: \sf \: (2x - 5)(3x + 3)$

<h3>The breadth of the rectangle is 3x + 3 </h3>

4 0

3 years ago

Find the discount for a stereo that is on sale for 12% off. The<br> original price is $200.00.

Fantom [35]

Answer:

$176

Step-by-step explanation:

$200 x 12%

12% --> 0.12

200 x 0.12 = 24 --> discounted off amount

200 - 24 = 176

3 0

2 years ago

Write the equation of the parabola in vertex form. Vertex (2,1), point (3, -4)

Ket [755]

Answer:

U=1-5(x-2)^2

Step-by-step explanation:

The equation of the parabola with vertex (h,k) is y=a(−h+x)2+k

Thus, the equation of the parabola is y=a(x−2)2+1

To find a, use the fact that the parabola passes through the point (3,−4): −4=a+1

Solving this equation, we get that a=−5

Thus, the equation of the parabola is y=1−5(x−2)^2

6 0

3 years ago

here are several types of objects. For each type of object, estimate how many there are in a stack that is 5 feet high. Be prepa

lorasvet [3.4K]

We need to estimate how tall is, on average, one of these object, and then count how many would be in a 5 feet high stack.

For example, on Google you may see that "1.5 cubic foot boxes are the standard box, manufactured by most companies". So, we assume that a standard cardboard box is 1.5 feet tall.

So, if we set the equation

$1.5k = 5 \iff k=\dfrac{5}{1.5}=3.\bar{3}$

So, there would be between 3 and 4 cardboard boxes in a 5 feet tall stack.

Similarly, we can see that the average book is 9 inches tall. 9 inches are 0.75

feet, so we have

$0.75k=5 \iff k=6.\bar{6}$

So, there would be between 6 and 7 books in a 5 feet tall stack.

The average brick is 75 millimeters tall, which means 0.25 feet tall. Again, we have

$0.25k=5 \iff k=20$

So, there would be 20 bricks in a 5 feet tall stack.

Finally, a coin is about 0.006 feet, which leads to

$0.006k=5 \iff k=833.\bar{3}$

So, there would be between 833 and 834 coins in a 5 feet tall stack.

5 0

3 years ago

Simplify 36x^4 42x^7

Alja [10]

The first step to simplify this equation is to multiply the numbers
1512 $x^{4}$ × $x^{7}$
next,, youll multiply the terms with the same base by adding their exponents
1512 $x^{4+7}$
finally,, youll add the numbers together
1512 $x^{11}$
since you cant simplify any further,, the correct answer to your question is 1512 $x^{11}$
let me know if you have any further questions
:)

4 0

2 years ago