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ankoles [38]
3 years ago
15

Please help quickly I only have 5 mins!!!

Mathematics
1 answer:
scZoUnD [109]3 years ago
6 0

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

Answer:

Let's go step-by-step ⇒

\frac{10x + 4}{7} = 8

= (\frac{7}{1})\frac{10x+4}{7} = \frac{8}{1}(\frac{7}{1})

= 10x+4=56

= 10x = 52

= x = 5.2

Your answer is x = 5.2

I hope this helps!

- sincerelynini

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Use the formula P=21+2w to find the width of a rectangle when the length is 18cm and perimeter is 48 ..
KIM [24]

Answer:

w=6

Step-by-step explanation:

P=2(L)+2(w)

48= 2(18)+2w

48=36+2w

12=2w

w=6

checking;

48= 2(18)+2(6)

48= 36+12

48= 48

8 0
3 years ago
Can someone help me on this problem i am stuck?
gavmur [86]
Not that hard. Its B

4 0
3 years ago
The odds of winning a teddy bear are 5:8. The probability of winning a framed picture is 43%. For which do you have a better cha
Serhud [2]

Answer:

I'd say you had a better chance going 5:8

4 0
3 years ago
a rectangle box has length 12 inches, width 15 inches, and a height of 17 inches. Find the angle between the diagonal of the box
ANTONII [103]

Answer:

0.7246 radians

Step-by-step explanation:

According to the Question,

Given that, a rectangle box has length 12 inches, width 15 inches, and a height of 17 inches

  • The length of the base diagonal (d) can be found using the Pythagorean theorem on length and width:

d = √{ (12)² +(15)² }  = √(144+225) = √369inches

  • The tangent of the angle is the ratio of the height of the box to this length

 Tan∅ = 17/√369

Taking the Tan^{-1} , we have

∅ = Tan^{-1}(17/√369) ≈ 0.7246 radians

4 0
3 years ago
What is the area of the largest rectangle that can be inscribed in an isosceles triangle with side lengths $8$, $\sqrt{80}$, and
e-lub [12.9K]
First let's solve the general case for an isosceles triangle of base 2 and height k. If the triangle is drawn so the base is on the x-axis and the apex is on the y-axis, the equation for the line containing the right side is
  y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
  area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
  d(area)/dw = 0 = -kw +k
  w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
  area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.

If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
  h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
  triangle area = (1/2)(8)(8) = 32

The maximum area of the inscribed rectangle will be half this value,
  16 square units

4 0
3 years ago
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