Ask question

Ask question

All categories

3 years ago

11

Janine wants a necklace 45 beads long

2 answers:

SCORPION-xisa [38]3 years ago

7 0

Answer:

9 red and 36 blue

Step-by-step explanation:

1+4=5 so 45/5=9

seraphim [82]3 years ago

4 0

If there are 45 beads, with a ratio of $1:4$
Then,
$\frac{45}{5}=9$
Then if $\frac{1}{4}=\frac{9}{x}$
Then $x=9*4$
Therefore, $x=36$
Meaning, there would be a ratio of 36 beads to 9 beads

You might be interested in

Which multiplication problem is modeled on the number line?<br> 2(– 4)<br> 4(2)<br> 2(4)<br> 4(– 2)

Sliva [168]

Answer:

4(-2)

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

disa [49]

Answer- 85,7 g
300:7= 42,85
42,85x2=85,7

7 0

3 years ago

St lies on the coordinate plane with s located at (3.2). The midpoint of dt is z(3,9) state the location of the endpoint T thank

77julia77 [94]

Answer:

T = (3,16)

Step-by-step explanation:

The midpoint formula is just another version of the pythagorean theorem, and it states that the midpoint between (x1,y1) and (x2,y2) is $\left(\frac{x_1+x_2}{2}\right),\:\left(\frac{y_1+y_2}{2}\right)$

Substituting what we know:

endpoint S is (3,2) and endpoint T is (x2,y2)

Midpoint Z is (3,9). We will apply the formula -- backwards.

$\left(\frac{3+x_2}{2}\right)=3$ , solving with algebra we get x2 = 3

So the x-coordinate of endpoint T is 3.

$\left(\frac{2+y_2}{2}\right)=9\quad$ , solving with algebra, we get y2 = 16.

So the y-coordinate of endpoint T is 16.

So the location of endpoint T is (3,16)

4 0

3 years ago

Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 & c, d are roots of x² - 20x + 2022 then the value of ac(a

Alja [10]

<h3><u>Correct Question :- </u></h3>

$\sf\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0 \: and \: \\ \sf \: c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0 \: then \:$

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =$

(a) 0

(b) 8000

(c) 8080

(d) 16000

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0}$

We know

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:ab = \dfrac{ - 2020}{1} = - 2020$

And

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:a + b = - \dfrac{20}{1} = - 20$

Also, given that

$\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0}$

$\rm \implies\:c + d = - \dfrac{( - 20)}{1} = 20$

and

$\rm \implies\:cd = \dfrac{2020}{1} = 2020$

Now, Consider

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)$

$\sf \: = {ca}^{2} - {ac}^{2} + {da}^{2} - {ad}^{2} + {cb}^{2} - {bc}^{2} + {db}^{2} - {bd}^{2}$

$\sf \: = {a}^{2}(c + d) + {b}^{2}(c + d) - {c}^{2}(a + b) - {d}^{2}(a + b)$

$\sf \: = (c + d)( {a}^{2} + {b}^{2}) - (a + b)( {c}^{2} + {d}^{2})$

$\sf \: = 20( {a}^{2} + {b}^{2}) + 20( {c}^{2} + {d}^{2})$

$\sf \: = 20\bigg[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\bigg]$

We know,

$\boxed{\tt{ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta \: }}$

So, using this, we get

$\sf \: = 20\bigg[ {(a + b)}^{2} - 2ab + {(c + d)}^{2} - 2cd\bigg]$

$\sf \: = 20\bigg[ {( - 20)}^{2} + 2(2020) + {(20)}^{2} - 2(2020)\bigg]$

$\sf \: = 20\bigg[ 400 + 400\bigg]$

$\sf \: = 20\bigg[ 800\bigg]$

$\sf \: = 16000$

Hence,

$\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}$

<em>So, option (d) is correct.</em>

4 0

2 years ago

15+(-11)+17=?<br><br> sjow how you got the answer

Tresset [83]

Answer:

21

Step-by-step explanation:

15 - 11 +17

plug in calculator

or 15-11 = 4

and

4 +17 = 21

8 0

3 years ago