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3 years ago

6

the sum of the digits 2014 is 7. How many whole numbers between 100 and 999 have 9 as the sum of their digits?

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Answer:

45

Step-by-step explanation:

this should be correct :)

There are 9 in the 100's

108

117

126

135

144

153

162

171

180

There are 8 in the 200's (starting to see a pattern?)

207

216

225

234

243

252

261

270

There are 7 in the 300's

306

315

324

333

342

351

360

9+8+7+6+5+4+3+2+1 = 45

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Does 14 and 10 have the same factor

iVinArrow [24]

Answer:

Yes

Step-by-step explanation:greatest common factor (GCF) of 10 and 14 is 2. We will now calculate the prime factors of 10 and 14, then find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 10 and 14.

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3 years ago

Find the circumference of the circle.

Masteriza [31]

Answer:

18.85, closest answer is A.

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3 years ago

HELP WITH ALL <br> Determine the composite functions of the following set of functions

elixir [45]

QUESTION 1

i) Given $f(x)=5x-2$ and $g(x)=x^2$ .

$(f\circ g)=f(g(x))$

$(f\circ g)=f(x^2)$

We plug in $x^2$ into $f(x)=5x-2$

$(f\circ g)=5x^2-2$

ii) Given $f(x)=5x-2$ and $g(x)=x^2$ .

$(g\circ f)=g(f(x))$

$(g\circ f)=g(5x-2)$

We plug in $5x-2$ into $g(x)=x^2$

$(g\circ f)=(5x-2)^2$

$(g\circ f)=25x^2-10x-10x+4$

$(g\circ f)=25x^2-20x+4$

QUESTION 2

i) Given $g(x)=x^2-4$ and $h(x)=\sqrt{2x+1}$ .

$(h\circ g)=h(g(x))$

$(h\circ g)=h(x^2-4)$

We plug in $x^2-4$ into $h(x)=\sqrt{2x+1}$

$(h\circ g)=\sqrt{2(x^2-4)+1}$

$(h\circ g)=\sqrt{2x^2-8+1}$

$(h\circ g)=\sqrt{2x^2-7}$

ii) Given $g(x)=x^2-4$ and $h(x)=\sqrt{2x+1}$ .

$(g\circ h)=g(h(x))$

$(g\circ h)=g(\sqrt{2x+1})$

We plug in $\sqrt{2x+1$ into $g(x)=x^2-4$

$(g\circ h)=(\sqrt{2x+1})^2-4$

$(g\circ h)=2x+1-4$

$(g\circ h)=2x-3$

QUESTION 3

i) Given $g(x)=x+5$ and $h(x)=\sqrt{x-4}$ .

$(g\circ h)=g(h(x))$

$(g\circ h)=g(\sqrt{x-4})$

We plug in $\sqrt{x-4}$ into $g(x)=x+5$

$(g\circ h)=\sqrt{(x-4)+5}$

ii) Given $g(x)=x+5$ and $h(x)=\sqrt{x-4}$ .

$(g\circ g)=g(g(x))$

$(g\circ g)=g(x+5)$

We plug in $x+5$ into $g(x)=x+5$

$(g\circ g)=(x+5)+5}$

$(g\circ g)=x+10}$

QUESTION 4

The given functions are;

$f(x)=-2x+1$ and $g(x)=-x^2$

$f\circ f=f(f(x))$

$f\circ f=f(-2x+1)$

$f\circ f=-2(-2x+1)+1)$

$f\circ f=4x-2+1)$

$f\circ f=4x-1)$

ii)

$g\circ g=g(g(x))$

$g\circ g=g(-x^2)$

$g\circ g=-(x^2)^2$

$g\circ g=-x^4$

QUESTION 5

$f(x)=5x-2;g(x)=x^2;h(x)=-3x$

$h\circ g\circ f=h\circ (g(f(x))$

$h\circ g\circ f=h\circ (g(5x-2))$

$h\circ g\circ f=h\circ ((5x-2)^2)$

$h\circ g\circ f=h\circ (25x^2-20x+4)$

$h\circ g\circ f=h(25x^2-20x+4)$

$h\circ g\circ f=-3(25x^2-20x+4)$

$h\circ g\circ f=-75x^2+60x-12)$

ii.

$f\circ f\circ g=f\circ (f(g(x))$

$f\circ f\circ g=f\circ (f(x^2)$

$f\circ f\circ g=f\circ (5x^2-2)$

$f\circ f\circ g=f(5x^2-2)$

$f\circ f\circ g=5(5x^2-2)-2$

$f\circ f\circ g=25x^2-10-2$

$f\circ f\circ g=25x^2-12$

7 0

3 years ago

What is the equation of the line that passes through (-1, 5) and (3, -7)?

enyata [817]

Answer:

$3x + y = 2$

Step-by-step explanation:

$y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

$y - 5 = \dfrac{-7 - 5}{3 - (-1)}(x - (-1))$

$y - 5 = \dfrac{-12}{4}(x + 1)$

$y - 5 = -3(x + 1)$

$y - 5 = -3x - 3$

$3x + y = 2$

3 0

3 years ago

What is the value of the expression –7 – 5 + 4?

vichka [17]

By solving this problem, I got D.-8.Right?

6 0

4 years ago