Ask question

Ask question

All categories

3 years ago

5

3. Reduce to lowest terms: 320 over 340

1 answer:

sineoko [7]3 years ago

5 0

320/340 can also be written as 16/17.

Divide/cancel both with 20.

You might be interested in

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

(4 + 2)(3^2) − 6 + 2 × 3 = 54<br> true<br> false

Pavlova-9 [17]

It would be true. (4+2)(3^2)-6+2*3=54 is true

Hope this helps :)

can you please make this the brainiest answer it would really help
<span />

8 0

4 years ago

The ratio of black marker in a box to blue markers in a box is 3:5. If there are 10 more blue markers than black markers,how man

fiasKO [112]

Answer:

$Total = 40$

Step-by-step explanation:

Given

$Black:Blue = 3:5$

$Blue = 10 + Black$

Required

Determine the total number of markers

First, we need to determine the ratio of black to blue such that

$Blue = 10 + Black$

$Black:Blue = 3:5$

Multiply the ratio by 5

$Black:Blue = 3 * 5 : 5 * 5$

$Black:Blue = 15 : 25$

By direct comparison:

$Black = 15$

$Blue = 25$

The above shows that:

$Blue = Black + 10$

So:

$Total = Black + Blue$

$Total = 25 + 15$

$Total = 40$

7 0

3 years ago

Lena ran 17 1/2 miles this week. She runs 2 1/2 miles each day.

eimsori [14]

2 1/2 * 7 = 17.5. she ran 7 days

5 0

4 years ago

Read 2 more answers

Jessica has $50 to spend on new clothes. The jacket that she wants to buy is $48.20. If the tax rate is 6% does Jessica have eno

Virty [35]

Answer:

Step-by-step explanation:

She has 50$. A jacket is on for 48.20$. The tax is 6% of the product you buy.

6% = 0.06

To find 6% you do:

0.06*48.20

Which is 2.892.

You ADD 2.892 to the price.

48.20+2.892=

51.092. Round it:

51.09 > 50.00

No, she does not have enough.

5 0

4 years ago