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kolbaska11 [484]
3 years ago
6

6^-7 • 6^-4 • 6^11 A 6^105 B 1 C 630 D 216

Mathematics
1 answer:
Nata [24]3 years ago
5 0

Answer:

Its 1

Step-by-step explanation:

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What is the domain and range of y=6
Ilya [14]

Answer:

domain -infinity

range(6)

Step-by-step explanation:

7 0
3 years ago
Catron evaluates the expression (negative 9) (2 and two-fifths) using the steps below.
Vlad [161]

Answer:

Catron's error is

"She did not follow order of operations"

Step-by-step explanation:

Catron evaluates the expression (negative 9) (2 and two-fifths)

That expression can be written as below

(-9)(2\frac{2}{5})

Catron's error is

"She did not follow order of operations"

The corrected steps are

Step1: Given expression is (-9)(2\frac{2}{5})

Step2: Convert mixed fraction into improper fraction

(-9)(2\frac{2}{5})=(-9)(\frac{12}{5})

Step3: Multiplying the terms

(-9)(2\frac{2}{5})=-7\frac{-108}{5}

Therefore solution (-9)(2\frac{2}{5})=-7\frac{-108}{5}

6 0
3 years ago
Read 2 more answers
3 - 8x = 8/5x written as a fraction
Ne4ueva [31]

Answer:

X=5/16                    

Explaination:

3 0
3 years ago
A ball is shot into the air using a brand new super-high-tech robotic arm to help baseball players practice catch fly balls.
stepladder [879]

Answer:

69 feet

Step-by-step explanation:

we have

h(t)=-16t^{2}+64t+5

where

h(t) is the height of the ball

t is the time in seconds

we know that the given equation is a vertical parabola open downward

The vertex is the maximum

so

the y-coordinate of the vertex represent the maximum height of the ball

Convert the quadratic equation into vertex form

The equation in vertex form is equal to

y=(x-h)^{2}+k

where

(h,k) is the vertex of the parabola

h(t)=-16t^{2}+64t+5

h(t)-5=-16t^{2}+64t

h(t)-5=-16(t^{2}-4t)

h(t)-5-64=-16(t^{2}-4t+4)

h(t)-69=-16(t^{2}-4t+4)

h(t)-69=-16(t-2)^{2}

h(t)=-16(t-2)^{2}+69

the vertex is the point (2,69)

therefore

The maximum height is 69 ft

6 0
3 years ago
A semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the
ira [324]

Answer:

  7 1/17

Step-by-step explanation:

A figure can be helpful.

The inscribed semicircle has its center at the midpoint of th base. It is tangent to the side of the isosceles triangle, so a radius makes a 90° angle there.

The long side of the isosceles triangle can be found from the Pythagorean theorem to be ...

  BC² = BD² +CD²

  BC² = 8² +15² = 289

  BC = √289 = 17

The radius mentioned (DE) creates right triangles that are similar to ∆BCD. In particular, we have ...

  (long side)/(hypotenuse) = DE/BD = CD/BC

  DE = BD·CD/BC = 8·15/17

  DE = 7 1/17 ≈ 7.059

8 0
3 years ago
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