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Delvig [45]
2 years ago
10

What is the solution to the equation below? 12 + square root 1 - 5x = 18

Mathematics
2 answers:
GrogVix [38]2 years ago
5 0

Answer:

-1

Step-by-step explanation:

Square root 1 is 1. Add that to the 12 to get 13. Subtract 13 from both sides to get: -5x = 5. Divide both sides by -5 to get your answer: -1

borishaifa [10]2 years ago
3 0

<u>Answer</u>:

x =  -7

<u>Explanation</u>:

  • 12 + \sqrt{1-5x} =18

  • \sqrt{1-5x}= 18-12

  • \sqrt{1-5x}= 6

  • 1-5x= 6^2

  • 1 -5x = 36

  • -5x  = 36-1

  • -5x =35

  • x =  -7
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A cabinet door has a perimeter of 62 inches its area is 228 in.² what are the dimensions of the door
Vlad1618 [11]
P=2(L+W)
A=LW


given
P=62
62=2(L+W)
divide 2
31=L+W
minus W
L=31-W

sub into other one
A=LW
A=(31-W)(W)
228=31W-W^2
times -1
W^2-31W=-228
add 228 both sides
W^2-31W+228=0
factor
what 2 numbers multiply to get 228 and add to get -31
-19 and -12
(W-19)(W-12)=0
set to zero

W-19=0
W=19

W-12=0
W=12

sub back

L=31-W
L=31-12
L=19
or
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the doorway is 12in by 19in

4 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
A wire is stretched from the ground to the top of an antenna tower. The wire is 15 feet long. The height of the tower is 3 feet
Mariana [72]

The distance d is 9 ft and the height is 12ft.

<h3>How to find the distance and the height?</h3>

Here we can model the situation with a right triangle, where the length of the wire is the hypotenuse.

The height is one cathetus and the distance is the other catheti.

Let's define:

  • h = height
  • d = distance.
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We know that the height of the tower is 3 ft larger than the distance, then:

h = d + 3ft

Now we can use the Pythagorean theorem, it says that the sum of the squares of the cathetus is equal to the square of the hypotenuse.

Then:

d^2 + (d + 3ft)^2 = (15ft)^2

Now we can solve this equation for d:

d^2 + d^2 + 6ft*d + 9ft^2 = (15ft)^2\\\\2d^2 + 6ft*d - 216 ft^2 = 0\\\\d^2 + 3ft*d - 108ft^2 = 0

Then the solutions are:

d = \frac{-3ft \pm \sqrt{(3ft)^2 - 4*(-108ft^2)} }{2} \\\\d = \frac{-3ft \pm 21ft }{2}

We only take the positive solution:

d = (-3ft + 21ft)/2 = 9ft

And the height is 3 ft more than that, so:

h = 9ft + 3ft = 12ft

The distance d is 9 ft and the height is 12ft.

If you want to learn more about right triangles:

brainly.com/question/2217700

#SPJ1

8 0
2 years ago
What's the circumference of a
Vlada [557]

Answer: C = 97.34 inches

Step-by-step explanation:

C = πd

= 3.14(31)

= 97.34

5 0
2 years ago
Please try to help me on all the questions not just one:)
nadezda [96]

Answer

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Step-by-step explanation:

im not sure about the others- hope this one helps a bit at least ;-;

5 0
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