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svetoff [14.1K]

3 years ago

5

5/7 of __ is 1/7 plz help asap

2 answers:

Alona [7]3 years ago

7 0

$\dfrac 57 x = \dfrac 17 \\\\\implies 5x =1 \\\\\implies x = \dfrac 15 \\\\\\\text{Hence }~ \dfrac 57 ~ \text{of}~ \dfrac 15 ~ \text{is }~\dfrac 17$

Vinvika [58]3 years ago

3 0

Answer:

1/5

Step-by-step explanation:

'of' is the same as multilpy

5/7 × ? = 1/7

? = 1/7 ÷ 5/7

? = 1/7 × 7/5 = 7/35

?=7/35

? = 1/5

check answer:

5/7 of 1/5 = 5/7 ×1/5 = 5/35 = 1/7

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Read 2 more answers

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

Jinny is selling bracelets for $3.50. So far, she has earned $266.00. She needs to earn more than $406.00 in order to meet her g

dem82 [27]

3.50x + 266 = 406
↓
40x
The answer is 40. She needs to sell 40 more bracelets to reach her goal.

8 0

3 years ago