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Roman55 [17]
2 years ago
12

5x5x5x5x6x7x8x8x9x9x9x9x9x9x10x12x13x14x15x16x17x18x19x20 WHAT DOES THIS EQUAL

Mathematics
2 answers:
Semenov [28]2 years ago
7 0
The answer your looking for is 5.441683e+23
trapecia [35]2 years ago
5 0

Answer:

5.441683e+23

Step-by-step explanation:

Hope it helped brainiest plz

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Need help w this please
rusak2 [61]
Solved using proportions.

8 0
3 years ago
What is the value of y?
NeX [460]

The value of y in the triangle is 2√5

The triangle above is a right angle triangle.

Characteristics of a right angle triangle:

  • One of its angle is equals to 90 degrees
  • The sides can be found using Pythagoras theorem.

Using Pythagoras theorem

  • c² = a² + b²

where

c = hypotenuse(longest side)

a and b are the 2 other legs.

Therefore,

6² = 4² + y²

36 = 16 + y²

y² = 36 - 16

y² = 20

y = √20

y = 2√5

learn more on right angle triangle here:brainly.com/question/3770177?referrer=searchResults

7 0
2 years ago
Read 2 more answers
Find the product <br> -1/2 y(2y3-8)
MissTica
0-(( \frac{1}{2} *y)*2y^3-8) \\ \\ then \ we \ would \ simplify \ 1/2 \\ \\ we \ factor y^3 - 4 \\ \\ 4 \ would \ not \ be \ a \ perfect \ cube. \\ \\ therefore, \ your \ answer \ would \ then \ be \ the \ following: \\ \\ \boxed{ -y * (y^3 - 4)}
8 0
3 years ago
Find the coordinates of the image of a triangle with vertices A(0, – 3), B(3, 0), and
lora16 [44]

Answer:

A'(-3,0), B'(0,-3) and C'(4,7)

Step-by-step explanation:

We are given that the vertices of triangle are A(0,-3), B(3,0) and C(-7,4).

We have to find the coordinates of the image of triangle under a rotation of 90° clockwise about the origin.

90° clockwise about the origin

Rule:(x,y)\rightarrow (y,-x)

Using the rule

The coordinates of A'

A(0,-3)\rightarrow A'(-3,0)

The coordinates of B'

B(3,0)\rightarrow B'(0,-3)

The coordinates of C'

C(-7,4)\rightarrow C'(4,7)

Hence, the vertices of image of triangle is given by

A'(-3,0), B'(0,-3) and C'(4,7)

8 0
2 years ago
Please Help !!Mr. Mudd gives each of his children $2000 to invest as part of a friendly family competition. The competition will
VikaD [51]

Answer:

Albert = $2159.07; Marie = $2244.99; Hans = $2188.35; Max = $2147.40

Marie is $10 000 richer

Step-by-step explanation:

Albert

(a) $1000 at 1.2 % compounded monthly

A = P\left(1 + \dfrac{r}{n}\right)^{nt}

A = 1000(1 + 0.001)¹²⁰ = $1127.43

(b) $500 losing 2%

0.98 × 500 = $490

(c) $500 compounded continuously at 0.8%

\begin{array}{rcl}A & = & Pe^{rt}\\& = & 500e^{0.008 \times 10}\\& = &\mathbf{\$541.64}\\\end{array}\\

(d) Balance

Total = 1127.43 + 490.00+ 541.64 = $2159.07

Marie

(a) 1500 at 1.4 % compounded quarterly

A = 1500(1 + 0.0035)⁴⁰ = $1724.99

(b) $500 gaining 4 %

1.04 × 500 = $520.00

(c) Balance

Total = 1724.99 + 520.00 = $2244.99

Hans

$2000 compounded continuously at 0.9 %

\begin{array}{rcl}A& = &2000e^{0.009 \times 10}\\& = &\mathbf{\$2188.35}\\\end{array}\\

Max

(a) $1000 decreasing exponentially at 0.5 % annually

A = 1000(1 - 0.005)¹⁰= $951.11

(b) $1000 at 1.8 % compounded biannually

A = 1000(1 + 0.009)²⁰ = $1196.29

(c) Balance

Total = 951.11 + 1196.29 = $2147.40

Marie is $ 10 000 richer at the end of the competition.

7 0
3 years ago
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