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3 years ago

6

Which division problem is modeled in the image? A) 2 ÷ 6 = 3 B) 2 ÷ 3 = 6 C) 3 ÷ 2 = 6 D) 6 ÷ 3 = 2 the image has 3 circles an 2

stars in each circle

2 answers:

irina1246 [14]3 years ago

8 0

Answer:

b

Step-by-step explanation:

iogann1982 [59]3 years ago

8 0

Answer:

its b

Step-by-step explanation:

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Given f(x)=x^2+8x+13. Enter the quadratic function in vertex form in the box f(x)= ___

Olenka [21]

Answer:

f(x) = (x + 4)² - 3

Step-by-step explanation:

The equation of a quadratic function in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Given

f(x) = x² + 8x + 13

To obtain f(x) in vertex form use the method of completing the square

add/ subtract (half the coefficient of the x- term)² to x² + 8x

f(x) = x² + 2(4)x + 16 - 16 + 13

f(x) = (x + 4)² - 3 ← in vertex form

8 0

3 years ago

A candle holder is in the shape of a triangular pyramid. The area of the base is 15 in2

EleoNora [17]

The volume of candle holder is 30 inches cubed.

Step-by-step explanation:

Given,

Area of base = A = 15 square inches

Height of candle holder = H = 6 inches

As the holder is in shape of triangular pyramid, therefore,

Volume of candle holder = $\frac{1}{3}AH$

Volume of candle holder = $\frac{1}{3}*15*6$

Volume of candle holder = $\frac{90}{3}=30\ in^3$

The volume of candle holder is 30 inches cubed.

7 0

3 years ago

Read 2 more answers

Consider the functions F and G in the tables below<br> PLEASE HELP PLEASE

d1i1m1o1n [39]

Answer:

Option A, As x increases, the rate of change exceeds the rate of f

7 0

3 years ago

5 POINTS PLEASE ANSWER!

Mnenie [13.5K]

Answer:

Value would decrease

Step-by-step explanation:

5 0

3 years ago

How do find the domain and range of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bx-8%7D%7D" id="TexFormula1" tit

Kisachek [45]

$\sqrt{x-8}$ is undefined if the argument $x-8$ is negative, so you first need to require that

$x-8\ge0\implies x\ge8$

We're not done yet, though, because $\dfrac1{\sqrt{x-8}}$ still doesn't exist when $x=8$ , so we remove this from the domain and we're left with $x>8$ , or in interval notation, $(8,\infty)$

To find the range, consider the limits of the function as you approach either endpoint of the domain.

$\displaystyle\lim_{x\to8^+}\frac1{\sqrt{x-8}}=+\infty$
$\displaystyle\lim_{x\to\infty}\frac1{\sqrt{x-8}}=0$

Since $\dfrac1{\sqrt{x-8}}$ is positive everywhere, the range is $(0,\infty)$

3 0

3 years ago