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4 years ago

You need 1 1/4 cups of sugar to make 20 cookiesto make 6 cookies, you will need ___ cups of sugar.

1 answer:

6 0

Use the rule of 3 simples

hope i understand it right that there is 1 integer and 1/4 cups of sugar or can being 11/4 cups too

1 integer 1/4 = 5/4

5/4 cups of sugar for 20 cookies
x -------------------- for 6 cookies
----------------------------------------------

x = 6*(5/4) / 20 = (30/4)/20 = 30/80 = 3/8 cups of sugar for 6 cookies

hope helped

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2. On a number line, G = 8 and H = –3. If H is the midpoint<br><br>of GI , find the coordinate of I.

MissTica

Answer:

-14

Step-by-step explanation:

the coordinate of I :

H = 8+I /2

-3 = 8+I /2

8+I = 2×(-3)

8+ I = -6

I = -6-8

I = -14

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3 years ago

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3 years ago

A quadratic equation ax^2+bx+c=0 has -11 and 4 as solutions. Find the values of b and c if the value of a is 1 (hint, use zero f

Alina [70]

Answer:

b = 7, c = -44

Step-by-step explanation:

If the quadratic equation has the solutions -11 and 4, the two factors are:

$(x+11)(x-4)=0$

Since when we use the zero factor property we get

x+11=0 ⇒ x= -11

x-4=0 ⇒ x=4

Thus, we have used the zero factor property in reverse to find the factorization of the quadratic equation.

Now we develop the multiplications between parenthesis:

$(x+11)(x-4)=0\\x^2-4x+11x-44=0\\x^2+7x-44=0$

So b is the number that accompanies the x: b = 7

and c is the independent number: c = -44

3 0

3 years ago

Uestion

Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}$

now, let's check the other path, JM and KL

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}$

so the red path will be $5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare$

3 0

2 years ago

Solve the equation using the quadratic formula. x^2-4x+3=0

almond37 [142]

Let's solve your equation step-by-step.

x2−4x+3=0

Step 1: Factor left side of equation.

(x−1)(x−3)=0

Step 2: Set factors equal to 0.
x-1=0 or x-3=0

For this one add one on both sides
x−1+1=0+1 x=1
For this one add 3 on both sides
x−3+3=0+3. x=3

x=1 or x=3

5 0

3 years ago